Today we will studyexponential equations.

Both elementary ones and those that are usually given on the Unified State Examination “for filling”.

Straight from the past Unified State Exam options.

However, after reading this article, all of them will become elementary for you.

Why?

Because you can follow step by step how I think when I solve them and learn to think the same way as I do.

Let's go!

What are exponential equations

If you have forgotten the following topics, then to get best result, Please, repeat:

1. Properties and
2. Solution and equations

Repeated? Amazing!

Then it will not be difficult for you to notice that the root of the equation is a number.

Do you understand exactly how I did it? Is it true? Then let's continue.

Now answer my question, what is equal to the third power? You're absolutely right: .

What power of two is eight? That's right - the third one! Because.

Well, now let's try to solve the following problem: Let me multiply the number by itself once and get the result.

The question is, how many times did I multiply by myself? You can of course check this directly:

\begin(align) & 2=2 \\ & 2\cdot 2=4 \\ & 2\cdot 2\cdot 2=8 \\ & 2\cdot 2\cdot 2\cdot 2=16 \\ \end( align)

Then you can conclude that I multiplied by myself times.

How else can you check this?

Here's how: directly by definition of degree: .

But, you must admit, if I asked how many times two needs to be multiplied by itself to get, say, you would tell me: I won’t fool myself and multiply by itself until I’m blue in the face.

And he would be absolutely right. Because how can you write down all the steps briefly(and brevity is the sister of talent)

where - these are the same ones "times", when you multiply by itself.

I think that you know (and if you don’t know, urgently, very urgently repeat the degrees!) that then my problem will be written in the form:

How can you reasonably conclude that:

So, unnoticed, I wrote down the simplest exponential equation:

And I even found him root. Don't you think that everything is completely trivial? I think exactly the same.

Here's another example for you:

But what to do?

After all, it cannot be written as a power of a (reasonable) number.

Let's not despair and note that both of these numbers are perfectly expressed through the power of the same number.

Then the original equation is transformed to the form:

Where, as you already understood, .

Let's not delay any longer and write it down definition:

In our case: .

These equations are solved by reducing them to the form:

followed by solving the equation

In fact, in the previous example we did just that: we got the following: And we solved the simplest equation.

It seems like nothing complicated, right? Let's practice on the simplest ones first examples:

We again see that the right and left sides of the equation need to be represented as powers of one number.

True, on the left this has already been done, but on the right there is a number.

But it’s okay, because my equation will miraculously transform into this:

What did I have to use here? What rule?

Rule of "degrees within degrees" which reads:

What if:

Before answering this question, let’s fill out the following table:

It is easy for us to notice that the smaller, the smaller the value, but nevertheless, all these values ​​are greater than zero.

AND IT WILL ALWAYS BE SO!!!

The same property is true FOR ANY BASIS WITH ANY INDICATOR!! (for any and).

Then what can we conclude about the equation?

Here's what it is: it has no roots! Just like any equation has no roots.

Now let's practice and Let's solve simple examples:

Let's check:

1. Here nothing will be required of you except knowledge of the properties of degrees (which, by the way, I asked you to repeat!)

As a rule, all lead to the smallest base: , .

Then the original equation will be equivalent to the following:

All I need is to use the properties of degrees:

When multiplying numbers with the same bases, the powers are added, and when dividing, they are subtracted.

Then I will get:

Well, now with a clear conscience I will move from the exponential equation to the linear one: \begin(align)
& 2x+1+2(x+2)-3x=5 \\
& 2x+1+2x+4-3x=5 \\
&x=0. \\
\end(align)

2. In the second example, we need to be more careful: the trouble is that on the left side we can’t possibly represent the same number as a power.

In this case it is sometimes useful represent numbers as a product of powers with different bases, but the same exponents:

The left side of the equation will take the form:

What did this give us?

Here's what: Numbers with different bases but the same exponents can be multiplied.In this case, the bases are multiplied, but the indicator does not change:

In my situation this will give:

\begin(align)
& 4\cdot ((64)^(x))((25)^(x))=6400,\\
& 4\cdot (((64\cdot 25))^(x))=6400,\\
& ((1600)^(x))=\frac(6400)(4), \\
& ((1600)^(x))=1600, \\
&x=1. \\
\end(align)

Not bad, right?

3. I don’t like it when, unnecessarily, I have two terms on one side of the equation and none on the other (sometimes, of course, this is justified, but now is not such a case).

I’ll move the minus term to the right:

Now, as before, I’ll write everything in terms of powers of three:

I add the degrees on the left and get an equivalent equation

You can easily find its root:

4. As in example three, the minus term has a place on the right side!

On my left, almost everything is fine, except for what?

Yes, the “wrong degree” of the two is bothering me. But I can easily fix this by writing: .

Eureka - on the left all the bases are different, but all the degrees are the same! Let's multiply immediately!

Here again, everything is clear: (if you don’t understand how I magically got the last equality, take a break for a minute, take a breath and read the properties of the degree again very carefully.

Who said you can skip a degree with a negative score? Well, that’s what I’m saying, no one). Now I will get:

\begin(align)
& ((2)^(4\left((x) -9 \right)))=((2)^(-1)) \\
& 4((x) -9)=-1 \\
& x=\frac(35)(4). \\
\end(align)

More exponential equations for practice

Here are some problems for you to practice, to which I will only give the answers (but in a “mixed” form). Solve them, check them, and you and I will continue our research!

Ready? Answers like this:

1. any number

Okay, okay, I was joking! Here are some sketches of solutions (some very brief!)

Don't you think it's no coincidence that one fraction on the left is the other one "inverted"? It would be a sin not to take advantage of this:

This rule is very often used when solving exponential equations, remember it well!

Then the original equation will become like this:

By solving this quadratic equation, you will get the following roots:

2. Another solution: dividing both sides of the equation by the expression on the left (or right).

Divide by what is on the right, then I get:

Where (why?!)

3. I don’t even want to repeat myself, everything has already been “chewed” so much.

4. equivalent to a quadratic equation, roots

5. You need to use the formula given in the first problem, then you will get that:

The equation has turned into a trivial identity that is true for any. Then the answer is any real number.

Well, now you have practiced solving simple exponential equations.

Real life examples of solving exponential equations

Now I want to give you a few life examples, which will help you understand why they are needed in principle.

Example 1 (mercantile)

Let you have rubles, but you want to turn it into rubles.

The bank offers you to take this money from you at an annual rate with monthly capitalization of interest (monthly accrual).

The question is, how many months do you need to open a deposit for to reach the required final amount?

Quite a mundane task, isn’t it?

Nevertheless, its solution is associated with the construction of the corresponding exponential equation: Let - the initial amount, - the final amount, - interest rate per period, - the number of periods.

In our case (if the rate is annual, then it is calculated per month).

Why is it divided by? If you don’t know the answer to this question, remember the topic “”!

Then we get this equation:

This exponential equation can only be solved using a calculator (its appearance hints at this, and this requires knowledge of logarithms, which we will get acquainted with a little later), which I will do: ...

Thus, to receive a million, we will need to make a deposit for a month (not very fast, right?).

Example 2 (regularly comes across on the Unified State Examination!! - the problem is taken from the “real” version)

During the collapse radioactive isotope its mass decreases according to the law, where (mg) is the initial mass of the isotope, (min.) is the time elapsed from the initial moment, (min.) is the half-life.

At the initial moment of time, the mass of the isotope is mg. Its half-life is min. After how many minutes will the mass of the isotope be equal to mg?

It’s okay: we just take and substitute all the data into the formula proposed to us:

Let's divide both parts by, "in the hope" that on the left we will get something digestible:

Well, we are very lucky! It’s on the left, then let’s move on to the equivalent equation:

Where is min.

As you can see, exponential equations have very real applications in practice.

Now I want to walk you through another (simple) way...

Solving exponential equations based on taking the common factor out of brackets and then grouping the terms.

Don't be scared by my words, you already came across this method in 7th grade when you studied polynomials. For example, if you needed:

Let's group: the first and third terms, as well as the second and fourth.

It is clear that the first and third are the difference of squares:

and the second and fourth have a common factor of three:

Then the original expression is equivalent to this:

Where to derive the common factor is no longer difficult:

Hence,

This is roughly what we will do when solving exponential equations: look for “commonality” among the terms and take it out of the brackets, and then - come what may, I believe that we will be lucky =))

Example No. 1

On the right is far from being a power of seven (I checked!) And on the left - it’s a little better, you can, of course, “chop off” the factor a from the second from the first term, and then deal with what you got, but let’s be more prudent with you.

I don't want to deal with the fractions that inevitably form when "selecting" , so shouldn't I rather take it out?

Then I won’t have any fractions: as they say, the wolves are fed and the sheep are safe:

Calculate the expression in brackets. Magically, magically, it turns out that (surprisingly, although what else should we expect?).

Then we reduce both sides of the equation by this factor. We get: , from.

Here's a more complicated example (quite a bit, really):

What a problem! We don't have one common ground here! It's not entirely clear what to do now. Let’s do what we can: first, move the “fours” to one side, and the “fives” to the other:

Now let's take out the "general" on the left and right:

So what now? What is the benefit of such a stupid group? At first glance it is not visible at all, but let's look deeper:

Well, now we’ll make sure that on the left we only have the expression c, and on the right - everything else. How do we do this? Here's how: Divide both sides of the equation first by (so we get rid of the exponent on the right), and then divide both sides by (so we get rid of the numeric factor on the left). Finally we get:

Incredible! On the left we have an expression, and on the right we have a simple expression.

Then we immediately conclude that

Example No. 2

I'll bring him short solution(without really bothering yourself with explanations), try to understand all the “subtleties” of the solution yourself.

Now for the final consolidation of the material covered. Try to solve the following problems yourself.

1. Let's take the common factor out of brackets: Where:
2. Let's present the first expression in the form: , divide both sides by and get that
3. , then the original equation is transformed to the form: Well, now a hint - look for where you and I have already solved this equation!
4. Imagine how, how, ah, well, then divide both sides by, so you get the simplest exponential equation.
5. Bring it out of the brackets.
6. Bring it out of the brackets.

EXPONENTARY EQUATIONS. MIDDLE LEVEL

I assume that after reading the first article, which talked about what are exponential equations and how to solve them, you have mastered the necessary minimum knowledge necessary to solve the simplest examples.

Now I will look at another method for solving exponential equations, this is...

Method for introducing a new variable (or replacement)

He solves most “difficult” problems on the topic of exponential equations (and not only equations).

This method is one of most frequently used in practice. First, I recommend that you familiarize yourself with the topic.

As you already understood from the name, the essence of this method is to introduce such a change of variable that your exponential equation will miraculously transform into one that you can easily solve.

All that remains for you after solving this very “simplified equation” is to make a “reverse replacement”: that is, return from the replaced to the replaced.

Let's illustrate what we just said with a very simple example:

Example 1. Simple replacement method

This equation can be solved using "simple replacement", as mathematicians disparagingly call it.

In fact, the replacement here is the most obvious. One has only to see that

Then the original equation will turn into this:

If we additionally imagine how, then it is absolutely clear what needs to be replaced: of course, . What then becomes the original equation? Here's what:

You can easily find its roots on your own: .

What should we do now?

It's time to return to the original variable.

What did I forget to mention? Namely: when replacing a certain degree with a new variable (that is, when replacing a type), I will be interested in only positive roots!

You yourself can easily answer why.

Thus, you and I are not interested, but the second root is quite suitable for us:

Then where from.

Answer:

As you can see, in the previous example, a replacement was just asking for our hands. Unfortunately, this is not always the case.

However, let’s not go straight to the sad stuff, but let’s practice with one more example with a fairly simple replacement

Example 2. Simple replacement method

It is clear that most likely it will have to be replaced (this is the smallest of the degrees included in our equation).

However, before introducing a replacement, our equation needs to be “prepared” for it, namely: , .

Then you can replace, as a result I get the following expression:

Oh horror: cubic equation with absolutely terrible formulas for its solution (well, speaking in general view). But let’s not despair right away, but let’s think about what we should do.

I'll suggest cheating: we know that to get a “beautiful” answer, we need to get it in the form of some power of three (why would that be, eh?).

Let's try to guess at least one root of our equation (I'll start guessing with powers of three).

First guess. Not a root. Alas and ah...

.
The left side is equal.
Right side: !

Eat! Guessed the first root. Now things will get easier!

Do you know about the “corner” division scheme? Of course you do, you use it when you divide one number by another. But few people know that the same can be done with polynomials.

There is one wonderful theorem:

Applying to my situation, this tells me that it is divisible without remainder by.

How is division carried out? Here's how:

I look at which monomial I should multiply by to get Clearly, then:

I subtract the resulting expression from, I get:

Now, what do I need to multiply by to get? It is clear that on, then I will get:

and again subtract the resulting expression from the remaining one:

Well, the last step is to multiply by and subtract from the remaining expression:

Hurray, division is over! What have we accumulated in private? Of course: .

Then we got the following expansion of the original polynomial:

Let's solve the second equation:

It has roots:

Then the original equation:

has three roots:

We will, of course, discard the last root, since it is less than zero. And the first two after reverse replacement will give us two roots:

Answer: ..

I didn’t mean to scare you with this example!

Rather, on the contrary, my goal was to show that although we had a fairly simple replacement, it nevertheless led to a rather complex equation, the solution of which required some special skills from us.

Well, no one is immune from this. But the replacement in this case was quite obvious.

Example #3 with a less obvious replacement:

It is not at all clear what we should do: the problem is that in our equation there are two different bases and one base cannot be obtained from the other by raising it to any (reasonable, naturally) power.

However, what do we see?

Both bases differ only in sign, and their product is the difference of squares equal to one:

Definition:

Thus, the numbers that are the bases in our example are conjugate.

In this case, the smart step would be multiply both sides of the equation by the conjugate number.

For example, on, then the left side of the equation will become equal to, and the right. If we make a substitution, then our original equation will become like this:

its roots, then, and remembering that, we get that.

Answer: , .

As a rule, the replacement method is sufficient to solve most “school” exponential equations.

Next tasks higher level difficulties are taken from the Unified State Exam options.

Problems of increased complexity from the Unified State Exam variants

You are already literate enough to solve these examples on your own. I will only give the required replacement.

1. Solve the equation:
2. Find the roots of the equation:
3. Solve the equation: . Find all the roots of this equation that belong to the segment:

And now some brief explanations and answers:

Equation #1.

Here it is enough for us to note that...

Then the original equation will be equivalent to this:

This equation can be solved by replacing

Do the further calculations yourself. In the end, your task will be reduced to solving simple trigonometric problems (depending on sine or cosine). We will look at solutions to similar examples in other sections.

Equation #2.

Here you can even do without substitution: just move the subtrahend to the right and represent both bases through powers of two: , and then go straight to the quadratic equation.

Equation #3

This is also solved in a fairly standard way: let’s imagine how.

Then, replacing, we get a quadratic equation: then,

You already know what a logarithm is, right? No? Then read the topic urgently!

The first root obviously does not belong to the segment, but the second one is unclear! But we will find out very soon! Since, then (this is a property of the logarithm!) Let’s compare:

Subtract from both sides, then we get:

The left side can be represented as:

multiply both sides by:

can be multiplied by, then

Then compare:

since then:

Then the second root belongs to the required interval

Answer:

As you can see, selection of roots of exponential equations requires a fairly deep knowledge of the properties of logarithms, so I advise you to be as careful as possible when solving exponential equations.

As you understand, in mathematics everything is interconnected! As my math teacher said: “mathematics, like history, cannot be read overnight.”

As a rule, all The difficulty in solving problems C1 is precisely the selection of the roots of the equation.

Another example for practice

It is clear that the equation itself is solved quite simply. By making a substitution, we reduce our original equation to the following:

First let's look at first root.

Let's compare and: since, then. (property of a logarithmic function, at).

Then it is clear that the first root does not belong to our interval.

Now the second root: . It is clear that (since the function at is increasing).

It remains to compare and...

since, then, at the same time.

This way I can “drive a peg” between the and.

This peg is a number. The first expression is less and the second is greater.

Then the second expression is greater than the first and the root belongs to the interval.

Answer: .

Finally, let's look at another example of an equation where the substitution is quite non-standard

An example of an equation with a non-standard substitution!

Let's start right away with what can be done, and what - in principle, can be done, but it is better not to do it.

You can imagine everything through the powers of three, two and six. What will this lead to?

It won’t lead to anything: a jumble of degrees, some of which will be quite difficult to get rid of.

What then is needed?

Let's notice that a

And what will this give us? And the fact that we can reduce the solution of this example to the solution of a fairly simple exponential equation!

First, let's rewrite our equation as:

Now let's divide both sides of the resulting equation by:

Eureka! Now we can replace, we get:

Well, now it’s your turn to solve exemplary problems, and I’ll only give them brief comments so that you don't go astray! Good luck!

1. The most difficult! It’s so hard to see a replacement here! But nevertheless, this example can be completely solved using highlighting a complete square. To solve it, it is enough to note that:

Then here's your replacement:

(Please note that here during our replacement we cannot discard the negative root!!! Why do you think?)

Now to solve the example you only have to solve two equations:

Both of them can be solved by a “standard replacement” (but the second one in one example!)

2. Notice that and make a replacement.

3. Decompose the number into coprime factors and simplify the resulting expression.

4. Divide the numerator and denominator of the fraction by (or, if you prefer) and make the substitution or.

5. Notice that the numbers and are conjugate.

SOLVING EXPONENTARY EQUATIONS USING THE LOGARITHMETHOD. ADVANCED LEVEL

In addition, let's look at another way - solving exponential equations using the logarithm method.

I can’t say that solving exponential equations using this method is very popular, but in some cases only it can lead us to the right decision our equation.

It is especially often used to solve the so-called “ mixed equations": that is, those where functions of different types occur.

For example, an equation of the form:

in the general case, it can only be solved by taking logarithms of both sides (for example, to the base), in which the original equation will turn into the following:

Let's look at the following example:

It is clear that according to the ODZ of the logarithmic function, we are only interested. However, this follows not only from the ODZ of the logarithm, but for one more reason. I think it won’t be difficult for you to guess which one it is.

Let's take the logarithm of both sides of our equation to the base:

As you can see, taking the logarithm of our original equation quickly led us to the correct (and beautiful!) answer.

Let's practice with one more example:

There’s nothing wrong here either: let’s take the logarithm of both sides of the equation to the base, then we get:

Let's make a replacement:

However, we missed something! Did you notice where I made a mistake? After all, then:

which does not satisfy the requirement (think where it came from!)

Answer:

Try to write down the solution to the exponential equations below:

Now compare your decision with this:

1. Let’s logarithm both sides to the base, taking into account that:

(the second root is not suitable for us due to replacement)

2. Logarithm to the base:

Let us transform the resulting expression to the following form:

EXPONENTARY EQUATIONS. BRIEF DESCRIPTION AND BASIC FORMULAS

Exponential equation

Equation of the form:

called the simplest exponential equation.

Properties of degrees

Approaches to solution

• Reduction to the same basis
• Reduction to the same exponent
• Variable replacement
• Simplifying the expression and applying one of the above.

Become a YouClever student,

Prepare for the Unified State Exam or Unified State Exam in mathematics,

And also get access to the YouClever textbook without restrictions...

On this lesson we will consider solving more complex exponential equations, recall the basic theoretical principles regarding exponential function.

1. Definition and properties of the exponential function, methods for solving the simplest exponential equations

Let us recall the definition and basic properties of the exponential function. The solution of all exponential equations and inequalities is based on these properties.

Exponential function is a function of the form , where the base is the degree and Here x is the independent variable, argument; y is the dependent variable, function.

Rice. 1. Graph of exponential function

The graph shows increasing and decreasing exponents, illustrating the exponential function with a base greater than one and less than one but greater than zero, respectively.

Both curves pass through the point (0;1)

Properties of the Exponential Function:

Scope: ;

Range of values: ;

The function is monotonic, increases with, decreases with.

A monotonic function takes each of its values ​​given a single argument value.

When the argument increases from minus to plus infinity, the function increases from zero inclusive to plus infinity. On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero, not inclusive.

2. Solving standard exponential equations

Let us remind you how to solve the simplest exponential equations. Their solution is based on the monotonicity of the exponential function. Almost all complex exponential equations can be reduced to such equations.

The equality of exponents with equal bases is due to the property of the exponential function, namely its monotonicity.

Solution method:

Equalize the bases of degrees;

Equate the exponents.

Let's move on to consider more complex exponential equations; our goal is to reduce each of them to the simplest.

Let's get rid of the root on the left side and bring the degrees to the same base:

In order to reduce a complex exponential equation to its simplest, substitution of variables is often used.

Let's use the power property:

We are introducing a replacement. Let it be then

Let's multiply the resulting equation by two and move all terms to the left side:

The first root does not satisfy the range of y values, so we discard it. We get:

Let's reduce the degrees to the same indicator:

Let's introduce a replacement:

Let it be then . With such a replacement, it is obvious that y takes on strictly positive values. We get:

We know how to solve such quadratic equations, we can write down the answer:

To make sure that the roots are found correctly, you can check using Vieta’s theorem, i.e., find the sum of the roots and their product and compare them with the corresponding coefficients of the equation.

We get:

3. Methodology for solving homogeneous exponential equations of the second degree

Let's study the following important type of exponential equations:

Equations of this type are called homogeneous second degrees with respect to the functions f and g. On the left side there is quadratic trinomial relative to f with parameter g or quadratic trinomial relative to g with parameter f.

Solution method:

This equation can be solved as a quadratic equation, but it is easier to do it differently. There are two cases to consider:

In the first case we get

In the second case, we have the right to divide by the highest degree and get:

It is necessary to introduce a change of variables, we obtain a quadratic equation for y:

Let us note that the functions f and g can be any, but we are interested in the case when these are exponential functions.

4. Examples of solving homogeneous equations

Let's move all the terms to the left side of the equation:

Since exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when:

We get:

Let's introduce a replacement: (according to the properties of the exponential function)

We got a quadratic equation:

We determine the roots using Vieta’s theorem:

The first root does not satisfy the range of values ​​of y, we discard it, we get:

Let's use the properties of degrees and reduce all degrees to simple bases:

It's easy to notice the functions f and g:

What is an exponential equation? Examples.

So, an exponential equation... A new unique exhibit in our general exhibition of a wide variety of equations!) As is almost always the case, the key word of any new mathematical term is the corresponding adjective that characterizes it. So it is here. The key word in the term “exponential equation” is the word "indicative". What does it mean? This word means that the unknown (x) is located in terms of any degrees. And only there! This is extremely important.

For example, these simple equations:

3 x +1 = 81

5 x + 5 x +2 = 130

4 2 2 x -17 2 x +4 = 0

Or even these monsters:

2 sin x = 0.5

Please immediately pay attention to one important thing: reasons degrees (bottom) – only numbers. But in indicators degrees (above) - a wide variety of expressions with an X. Absolutely any.) Everything depends on the specific equation. If, suddenly, x appears somewhere else in the equation, in addition to the indicator (say, 3 x = 18 + x 2), then such an equation will already be an equation mixed type. Such equations do not have clear rules for solving them. Therefore, we will not consider them in this lesson. To the delight of the students.) Here we will consider only exponential equations in their “pure” form.

Generally speaking, not all and not always even pure exponential equations can be solved clearly. But among all the rich variety of exponential equations, there are certain types that can and should be solved. It is these types of equations that we will consider. And we’ll definitely solve the examples.) So let’s get comfortable and off we go! As in computer “shooters”, our journey will take place through levels.) From elementary to simple, from simple to average and from average to complex. Along the way, a secret level will also await you - techniques and methods for solving non-standard examples. The ones you don't read about most school textbooks... Well, at the end, of course, the final boss awaits you in the form of homework.)

Level 0. What is the simplest exponential equation? Solving simple exponential equations.

First, let's look at some frank elementary stuff. You have to start somewhere, right? For example, this equation:

2 x = 2 2

Even without any theories, by simple logic and common sense it is clear that x = 2. There is no other way, right? No other meaning of X is suitable... And now let’s turn our attention to record of decision this cool exponential equation:

2 x = 2 2

X = 2

What happened to us? And the following happened. We actually took it and... simply threw out the same bases (twos)! Completely thrown out. And, the good news is, we hit the bull’s eye!

Yes, indeed, if in an exponential equation there are left and right identical numbers in any powers, then these numbers can be discarded and simply equate the exponents. Mathematics allows.) And then you can work separately with the indicators and solve a much simpler equation. Great, right?

Here is the key idea for solving any (yes, exactly any!) exponential equation: using identical transformations, it is necessary to ensure that the left and right sides of the equation are identical base numbers in various powers. And then you can safely remove the same bases and equate the exponents. And work with a simpler equation.

Now let’s remember the iron rule: it is possible to remove identical bases if and only if the numbers on the left and right of the equation have base numbers in splendid isolation.

What does it mean, in splendid isolation? This means without any neighbors and coefficients. Let me explain.

For example, in Eq.

3 3 x-5 = 3 2 x +1

Threes cannot be removed! Why? Because on the left we have not just a lonely three to the degree, but work 3·3 x-5 . An extra three interferes: the coefficient, you understand.)

The same can be said about the equation

5 3 x = 5 2 x +5 x

Here, too, all the bases are the same - five. But on the right we don’t have a single power of five: there is a sum of powers!

In short, we have the right to remove identical bases only when our exponential equation looks like this and only like this:

af (x) = a g (x)

This type of exponential equation is called the simplest. Or, scientifically, canonical . And no matter what convoluted equation we have in front of us, we will, one way or another, reduce it to precisely this simplest (canonical) form. Or, in some cases, to totality equations of this kind. Then our simplest equation can be rewritten in general form like this:

F(x) = g(x)

That's all. This would be an equivalent conversion. In this case, f(x) and g(x) can be absolutely any expressions with an x. Whatever.

Perhaps a particularly inquisitive student will wonder: why on earth do we so easily and simply discard the same bases on the left and right and equate the exponents? Intuition is intuition, but what if, in some equation and for some reason, this approach turns out to be incorrect? Is it always legal to throw out the same grounds? Unfortunately, for a rigorous mathematical answer to this interesting question you need to dive quite deeply and seriously into general theory device and function behavior. And a little more specifically - in the phenomenon strict monotony. In particular, strict monotony exponential functiony= a x. Since it is the exponential function and its properties that underlie the solution of exponential equations, yes.) A detailed answer to this question will be given in a separate special lesson dedicated to solving complex non-standard equations using the monotonicity of different functions.)

Explaining this point in detail now would only blow the minds of the average schoolchild and scare him away ahead of time with a dry and heavy theory. I won’t do this.) Because our main task at the moment is learn to solve exponential equations! The simplest ones! Therefore, let’s not worry yet and boldly throw out the same reasons. This Can, take my word for it!) And then we solve the equivalent equation f(x) = g(x). As a rule, simpler than the original exponential.

It is assumed, of course, that people already know how to solve at least , and equations, without x’s in exponents.) For those who still don’t know how, feel free to close this page, follow the relevant links and fill in the old gaps. Otherwise you will have a hard time, yes...

I'm not talking about irrational, trigonometric and other brutal equations that can also emerge in the process of eliminating the foundations. But don’t be alarmed, we won’t consider outright cruelty in terms of degrees for now: it’s too early. We will train only on the most simple equations.)

Now let's look at equations that require some additional effort to reduce them to the simplest. For the sake of distinction, let's call them simple exponential equations. So, let's move to the next level!

Level 1. Simple exponential equations. Let's recognize the degrees! Natural indicators.

The key rules in solving any exponential equations are rules for dealing with degrees. Without this knowledge and skills nothing will work. Alas. So, if there are problems with the degrees, then first you are welcome. In addition, we will also need . These transformations (two of them!) are the basis for solving all mathematical equations in general. And not only demonstrative ones. So, whoever forgot, also take a look at the link: I don’t just put them there.

But operations with powers and identity transformations alone are not enough. Personal observation and ingenuity are also required. We need the same reasons, don't we? So we examine the example and look for them in an explicit or disguised form!

For example, this equation:

3 2 x – 27 x +2 = 0

First look at grounds. They are... different! Three and twenty seven. But it’s too early to panic and despair. It's time to remember that

27 = 3 3

Numbers 3 and 27 are relatives by degree! And close ones.) Therefore, we have every right to write:

27 x +2 = (3 3) x+2

Now let’s connect our knowledge about actions with degrees(and I warned you!). There is a very useful formula there:

(a m) n = a mn

If you now put it into action, it works out great:

27 x +2 = (3 3) x+2 = 3 3(x +2)

The original example now looks like this:

3 2 x – 3 3(x +2) = 0

Great, the bases of the degrees have leveled out. That's what we wanted. Half the battle is done.) Now we launch the basic identity transformation - move 3 3(x +2) to the right. No one has canceled the elementary operations of mathematics, yes.) We get:

3 2 x = 3 3(x +2)

What does this type of equation give us? And the fact that now our equation is reduced to canonical form: on the left and right there are the same numbers (threes) in powers. Moreover, both three are in splendid isolation. Feel free to remove the triples and get:

2x = 3(x+2)

We solve this and get:

X = -6

That's it. This is the correct answer.)

Now let’s think about the solution. What saved us in this example? Knowledge of the powers of three saved us. How exactly? We identified number 27 contains an encrypted three! This trick (encoding the same base under different numbers) is one of the most popular in exponential equations! Unless it's the most popular. Yes, and in the same way, by the way. This is why observation and the ability to recognize powers of other numbers in numbers are so important in exponential equations!

Practical advice:

You need to know the powers of popular numbers. In the face!

Of course, anyone can raise two to the seventh power or three to the fifth power. Not in my mind, but at least in a draft. But in exponential equations, much more often it is not necessary to raise to a power, but, on the contrary, to find out what number and to what power is hidden behind the number, say, 128 or 243. And this is more complicated than simple raising, you will agree. Feel the difference, as they say!

Since the ability to recognize degrees in person will be useful not only at this level, but also at the next ones, here’s a small task for you:

Determine what powers and what numbers the numbers are:

4; 8; 16; 27; 32; 36; 49; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729; 1024.

Answers (randomly, of course):

27 2 ; 2 10 ; 3 6 ; 7 2 ; 2 6 ; 9 2 ; 3 4 ; 4 3 ; 10 2 ; 2 5 ; 3 5 ; 7 3 ; 16 2 ; 2 7 ; 5 3 ; 2 8 ; 6 2 ; 3 3 ; 2 9 ; 2 4 ; 2 2 ; 4 5 ; 25 2 ; 4 4 ; 6 3 ; 8 2 ; 9 3 .

Yes, yes! Don't be surprised that there are more answers than tasks. For example, 2 8, 4 4 and 16 2 are all 256.

Level 2. Simple exponential equations. Let's recognize the degrees! Negative and fractional indicators.

At this level we are already using our knowledge of degrees to the fullest. Namely, we involve negative and fractional indicators in this fascinating process! Yes, yes! We need to increase our power, right?

For example, this terrible equation:

Again, the first glance is at the foundations. The reasons are different! And this time they are not even remotely similar to each other! 5 and 0.04... And to eliminate the bases, the same ones are needed... What to do?

It's OK! In fact, everything is the same, it’s just that the connection between the five and 0.04 is visually poorly visible. How can we get out? Let's move on to the number 0.04 as an ordinary fraction! And then, you see, everything will work out.)

0,04 = 4/100 = 1/25

Wow! It turns out that 0.04 is 1/25! Well, who would have thought!)

So how? Is it now easier to see the connection between the numbers 5 and 1/25? That's it...

And now according to the rules of actions with degrees with negative indicator You can write with a steady hand:

That's great. So we got to the same base - five. Now we replace the inconvenient number 0.04 in the equation with 5 -2 and get:

Again, according to the rules of operations with degrees, we can now write:

(5 -2) x -1 = 5 -2(x -1)

Just in case, I remind you (in case anyone doesn’t know) that basic rules actions with powers are valid for any indicators! Including for negative ones.) So, feel free to take and multiply the indicators (-2) and (x-1) according to the appropriate rule. Our equation gets better and better:

All! Apart from lonely fives, there is nothing else in the powers on the left and right. The equation is reduced to canonical form. And then - along the knurled track. We remove the fives and equate the indicators:

x 2 –6 x+5=-2(x-1)

The example is almost solved. All that's left is elementary middle school math - open (correctly!) the brackets and collect everything on the left:

x 2 –6 x+5 = -2 x+2

x 2 –4 x+3 = 0

We solve this and get two roots:

x 1 = 1; x 2 = 3

That's all.)

Now let's think again. In this example, we again had to recognize the same number in different degrees! Namely, to see an encrypted five in the number 0.04. And this time - in negative degree! How did we do this? Right off the bat - no way. But after moving from the decimal fraction 0.04 to the common fraction 1/25, everything became clear! And then the whole decision went like clockwork.)

Therefore, another green practical advice.

If the exponential equation contains decimal fractions, then we move from decimals to ordinary ones. IN ordinary fractions It's much easier to recognize powers of many popular numbers! After recognition, we move from fractions to powers with negative exponents.

Keep in mind that this trick occurs very, very often in exponential equations! But the person is not in the subject. He looks, for example, at the numbers 32 and 0.125 and gets upset. Unbeknownst to him, this is one and the same two, only in different degrees... But you’re already in the know!)

Solve the equation:

In! It looks like quiet horror... However, appearances are deceiving. This is the simplest exponential equation, despite its intimidating appearance. And now I will show it to you.)

First, let’s look at all the numbers in the bases and coefficients. They are, of course, different, yes. But we will still take a risk and try to make them identical! Let's try to get to the same number in different powers. Moreover, preferably, the numbers are as small as possible. So, let's start decoding!

Well, with the four everything is immediately clear - it’s 2 2. So, that's something already.)

With a fraction of 0.25 – it’s still unclear. Need to check. Let's use practical advice - move from a decimal fraction to an ordinary fraction:

0,25 = 25/100 = 1/4

Much better already. Because now it is clearly visible that 1/4 is 2 -2. Great, and the number 0.25 is also akin to two.)

So far so good. But the worst number of all remains - square root of two! What to do with this pepper? Can it also be represented as a power of two? And who knows...

Well, let's dive into our treasury of knowledge about degrees again! This time we additionally connect our knowledge about roots. From the 9th grade course, you and I should have learned that any root, if desired, can always be turned into a degree with a fractional indicator.

Like this:

In our case:

Wow! It turns out that the square root of two is 2 1/2. That's it!

That's great! All our inconvenient numbers actually turned out to be an encrypted two.) I don’t argue, somewhere very sophisticatedly encrypted. But we are also improving our professionalism in solving such ciphers! And then everything is already obvious. In our equation we replace the numbers 4, 0.25 and the root of two by powers of two:

All! The bases of all degrees in the example became the same - two. And now standard actions with degrees are used:

a ma n = a m + n

a m:a n = a m-n

(a m) n = a mn

For the left side you get:

2 -2 ·(2 2) 5 x -16 = 2 -2+2(5 x -16)

For the right side it will be:

And now our evil equation looks like this:

For those who haven’t figured out exactly how this equation came about, then the question here is not about exponential equations. The question is about actions with degrees. I asked you to urgently repeat it to those who have problems!

Here is the finish line! Received canonical view exponential equation! So how? Have I convinced you that everything is not so scary? ;) We remove the twos and equate the indicators:

All that remains is to solve this linear equation. How? With the help of identical transformations, of course.) Decide what’s going on! Multiply both sides by two (to remove the fraction 3/2), move the terms with X's to the left, without X's to the right, bring similar ones, count - and you will be happy!

Everything should turn out beautifully:

X=4

Now let’s think about the solution again. In this example, we were helped by the transition from square root To degree with exponent 1/2. Moreover, only such a cunning transformation helped us reach the same base (two) everywhere, which saved the situation! And, if not for it, then we would have every chance to freeze forever and never cope with this example, yes...

Therefore, we do not neglect the next practical advice:

If an exponential equation contains roots, then we move from roots to powers with fractional exponents. Very often only such a transformation clarifies the further situation.

Of course, negative and fractional powers are much more complicated natural degrees. At least from the point of view of visual perception and, especially, recognition from right to left!

It is clear that directly raising, for example, two to the power of -3 or four to the power of -3/2 is not such a big problem. For those in the know.)

But go, for example, immediately realize that

0,125 = 2 -3

Or

Here, only practice and rich experience rule, yes. And, of course, a clear idea, What is a negative and fractional degree? And also - practical advice! Yes, yes, those same ones green.) I hope that they will still help you better navigate the entire diverse variety of degrees and significantly increase your chances of success! So let's not neglect them. It’s not for nothing that I sometimes write in green.)

But if you get to know each other even with such exotic powers as negative and fractional ones, then your capabilities in solving exponential equations will expand enormously, and you will be able to handle almost any type of exponential equations. Well, if not any, then 80 percent of all exponential equations - for sure! Yes, yes, I'm not joking!

So, our first part of our introduction to exponential equations has come to its logical conclusion. And, as an intermediate workout, I traditionally suggest doing a little self-reflection.)

Task 1.

So that my words about deciphering negative and fractional powers not in vain, I suggest you play a little game!

Express numbers as powers of two:

Answers (in disarray):

Did it work? Great! Then we do a combat mission - we solve the simplest and simplest exponential equations!

Task 2.

Solve the equations (all answers are a mess!):

5 2x-8 = 25

2 5x-4 – 16 x+3 = 0

Answers:

x = 16

x 1 = -1; x 2 = 2

x = 5

Did it work? Indeed, it’s much simpler!

Then we solve the next game:

(2 x +4) x -3 = 0.5 x 4 x -4

35 1-x = 0.2 - x ·7 x

Answers:

x 1 = -2; x 2 = 2

x = 0,5

x 1 = 3; x 2 = 5

And these examples are one left? Great! You are growing! Then here are some more examples for you to snack on:

Answers:

x = 6

x = 13/31

x = -0,75

x 1 = 1; x 2 = 8/3

And is this decided? Well, respect! I take my hat off.) This means that the lesson was not in vain, and the initial level of solving exponential equations can be considered successfully mastered. Next levels and more complex equations are ahead! And new techniques and approaches. And non-standard examples. And new surprises.) All this is in the next lesson!

Did something go wrong? This means that most likely the problems are in . Or in . Or both at once. I'm powerless here. I can once again suggest only one thing - don’t be lazy and follow the links.)

To be continued.)

This lesson is intended for those who are just beginning to learn exponential equations. As always, let's start with the definition and simple examples.

If you are reading this lesson, then I suspect that you already have at least a minimal understanding of the simplest equations - linear and quadratic: $56x-11=0$; $((x)^(2))+5x+4=0$; $((x)^(2))-12x+32=0$, etc. Being able to solve such constructions is absolutely necessary in order not to “get stuck” in the topic that will now be discussed.

So, exponential equations. Let me give you a couple of examples:

\[((2)^(x))=4;\quad ((5)^(2x-3))=\frac(1)(25);\quad ((9)^(x))=- 3\]

Some of them may seem more complex to you, while others, on the contrary, are too simple. But they all have one important feature in common: their notation contains the exponential function $f\left(x \right)=((a)^(x))$. Thus, let's introduce the definition:

An exponential equation is any equation containing an exponential function, i.e. expression of the form $((a)^(x))$. In addition to the specified function, such equations can contain any other algebraic constructions - polynomials, roots, trigonometry, logarithms, etc.

OK then. We've sorted out the definition. Now the question is: how to solve all this crap? The answer is both simple and complex.

Let's start with the good news: from my experience of teaching many students, I can say that most of them find exponential equations much easier than the same logarithms, and even more so trigonometry.

But there is bad news: sometimes the compilers of problems for all kinds of textbooks and exams are struck by “inspiration”, and their drug-inflamed brain begins to produce such brutal equations that solving them becomes problematic not only for students - even many teachers get stuck on such problems.

However, let's not talk about sad things. And let's return to those three equations that were given at the very beginning of the story. Let's try to solve each of them.

First equation: $((2)^(x))=4$. Well, to what power must you raise the number 2 to get the number 4? Probably the second? After all, $((2)^(2))=2\cdot 2=4$ - and we got the correct numerical equality, i.e. indeed $x=2$. Well, thanks, Cap, but this equation was so simple that even my cat could solve it. :)

Let's look at the following equation:

\[((5)^(2x-3))=\frac(1)(25)\]

But here it’s a little more complicated. Many students know that $((5)^(2))=25$ is the multiplication table. Some also suspect that $((5)^(-1))=\frac(1)(5)$ is essentially the definition of negative powers (similar to the formula $((a)^(-n))= \frac(1)(((a)^(n)))$).

Finally, only a select few realize that these facts can be combined and yield the following result:

\[\frac(1)(25)=\frac(1)(((5)^(2)))=((5)^(-2))\]

Thus, our original equation will be rewritten as follows:

\[((5)^(2x-3))=\frac(1)(25)\Rightarrow ((5)^(2x-3))=((5)^(-2))\]

But this is already completely solvable! On the left in the equation there is an exponential function, on the right in the equation there is an exponential function, there is nothing else anywhere except them. Therefore, we can “discard” the bases and stupidly equate the indicators:

We have obtained the simplest linear equation that any student can solve in just a couple of lines. Okay, in four lines:

\[\begin(align)& 2x-3=-2 \\& 2x=3-2 \\& 2x=1 \\& x=\frac(1)(2) \\\end(align)\]

If you don’t understand what was happening in the last four lines, be sure to return to the topic “ linear equations"and repeat it. Because without a clear understanding of this topic, it is too early for you to take on exponential equations.

\[((9)^(x))=-3\]

So how can we solve this? First thought: $9=3\cdot 3=((3)^(2))$, so the original equation can be rewritten as follows:

\[((\left(((3)^(2)) \right))^(x))=-3\]

Then we remember that when raising a power to a power, the exponents are multiplied:

\[((\left(((3)^(2)) \right))^(x))=((3)^(2x))\Rightarrow ((3)^(2x))=-(( 3)^(1))\]

\[\begin(align)& 2x=-1 \\& x=-\frac(1)(2) \\\end(align)\]

And for such a decision we will receive a honestly deserved two. For, with the equanimity of a Pokemon, we sent the minus sign in front of the three to the power of this very three. But you can’t do that. And here's why. Take a look at the different powers of three:

\[\begin(matrix) ((3)^(1))=3& ((3)^(-1))=\frac(1)(3)& ((3)^(\frac(1)( 2)))=\sqrt(3) \\ ((3)^(2))=9& ((3)^(-2))=\frac(1)(9)& ((3)^(\ frac(1)(3)))=\sqrt(3) \\ ((3)^(3))=27& ((3)^(-3))=\frac(1)(27)& (( 3)^(-\frac(1)(2)))=\frac(1)(\sqrt(3)) \\\end(matrix)\]

When compiling this tablet, I did not pervert as much as possible: I considered positive degrees, and negative ones, and even fractional ones... well, where is there at least one negative number? He's gone! And it cannot be, because the exponential function $y=((a)^(x))$, firstly, always takes only positive values ​​(no matter how much one is multiplied or divided by two, it will still be a positive number), and secondly, the base of such a function - the number $a$ - is by definition a positive number!

Well, how then to solve the equation $((9)^(x))=-3$? But no way: there are no roots. And in this sense, exponential equations are very similar to quadratic equations - there may also be no roots. But if in quadratic equations the number of roots is determined by the discriminant (positive discriminant - 2 roots, negative - no roots), then in exponential equations everything depends on what is to the right of the equal sign.

Thus, we formulate the key conclusion: the simplest exponential equation of the form $((a)^(x))=b$ has a root if and only if $b \gt 0$. Knowing this simple fact, you can easily determine whether the equation proposed to you has roots or not. Those. Is it worth solving it at all or immediately writing down that there are no roots.

This knowledge will help us many times when we have to decide more complex tasks. For now, enough of the lyrics - it’s time to study the basic algorithm for solving exponential equations.

How to Solve Exponential Equations

So, let's formulate the problem. It is necessary to solve the exponential equation:

\[((a)^(x))=b,\quad a,b \gt 0\]

According to the “naive” algorithm that we used earlier, it is necessary to represent the number $b$ as a power of the number $a$:

In addition, if instead of the variable $x$ there is any expression, we will get a new equation that can already be solved. For example:

\[\begin(align)& ((2)^(x))=8\Rightarrow ((2)^(x))=((2)^(3))\Rightarrow x=3; \\& ((3)^(-x))=81\Rightarrow ((3)^(-x))=((3)^(4))\Rightarrow -x=4\Rightarrow x=-4; \\& ((5)^(2x))=125\Rightarrow ((5)^(2x))=((5)^(3))\Rightarrow 2x=3\Rightarrow x=\frac(3)( 2). \\\end(align)\]

And oddly enough, this scheme works in about 90% of cases. What then about the remaining 10%? The remaining 10% are slightly “schizophrenic” exponential equations of the form:

\[((2)^(x))=3;\quad ((5)^(x))=15;\quad ((4)^(2x))=11\]

Well, to what power do you need to raise 2 to get 3? First? But no: $((2)^(1))=2$ is not enough. Second? No either: $((2)^(2))=4$ is too much. Which one then?

Knowledgeable students have probably already guessed: in such cases, when it is not possible to solve it “beautifully”, the “heavy artillery” - logarithms - comes into play. Let me remind you that using logarithms, any positive number can be represented as a power of any other positive number(except for one):

Remember this formula? When I tell my students about logarithms, I always warn: this formula (it is also the main logarithmic identity or, if you like, the definition of a logarithm) will haunt you for a very long time and “pop up” in the most unexpected places. Well, she surfaced. Let's look at our equation and this formula:

\[\begin(align)& ((2)^(x))=3 \\& a=((b)^(((\log )_(b))a)) \\\end(align) \]

If we assume that $a=3$ is our original number on the right, and $b=2$ is the very base of the exponential function to which we so want to reduce the right-hand side, we get the following:

\[\begin(align)& a=((b)^(((\log )_(b))a))\Rightarrow 3=((2)^(((\log )_(2))3 )); \\& ((2)^(x))=3\Rightarrow ((2)^(x))=((2)^(((\log )_(2))3))\Rightarrow x=( (\log )_(2))3. \\\end(align)\]

We received a slightly strange answer: $x=((\log )_(2))3$. In some other task, many would have doubts with such an answer and would begin to double-check their solution: what if an error had crept in somewhere? I hasten to please you: there is no error here, and logarithms in the roots of exponential equations are a completely typical situation. So get used to it. :)

Now let’s solve the remaining two equations by analogy:

\[\begin(align)& ((5)^(x))=15\Rightarrow ((5)^(x))=((5)^(((\log )_(5))15)) \Rightarrow x=((\log )_(5))15; \\& ((4)^(2x))=11\Rightarrow ((4)^(2x))=((4)^(((\log )_(4))11))\Rightarrow 2x=( (\log )_(4))11\Rightarrow x=\frac(1)(2)((\log )_(4))11. \\\end(align)\]

That's it! By the way, the last answer can be written differently:

We introduced a multiplier to the argument of the logarithm. But no one is stopping us from adding this factor to the base:

Moreover, all three options are correct - it’s simple different shapes records of the same number. Which one to choose and write down in this solution is up to you to decide.

Thus, we have learned to solve any exponential equations of the form $((a)^(x))=b$, where the numbers $a$ and $b$ are strictly positive. However, the harsh reality of our world is that such simple tasks will be encountered very, very rarely. More often than not you will come across something like this:

\[\begin(align)& ((4)^(x))+((4)^(x-1))=((4)^(x+1))-11; \\& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2.7)^(1-x))=0.09. \\\end(align)\]

So how can we solve this? Can this be solved at all? And if so, how?

Don't panic. All these equations quickly and easily reduce to the simple formulas that we have already considered. You just need to remember a couple of tricks from the algebra course. And of course, there are no rules for working with degrees. I'll tell you about all this now. :)

Converting Exponential Equations

The first thing to remember: any exponential equation, no matter how complex it may be, one way or another must be reduced to the simplest equations - the ones that we have already considered and which we know how to solve. In other words, the scheme for solving any exponential equation looks like this:

1. Write down the original equation. For example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
2. Do some weird shit. Or even some crap called "convert an equation";
3. At the output, get the simplest expressions of the form $((4)^(x))=4$ or something else like that. Moreover, one initial equation can give several such expressions at once.

Everything is clear with the first point - even my cat can write the equation on a piece of paper. The third point also seems to be more or less clear - we have already solved a whole bunch of such equations above.

But what about the second point? What kind of transformations? Convert what into what? And how?

Well, let's find out. First of all, I would like to note the following. All exponential equations are divided into two types:

1. The equation is composed of exponential functions with the same base. Example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
2. The formula contains exponential functions with different bases. Examples: $((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x))$ and $((100)^(x-1) )\cdot ((2,7)^(1-x))=$0.09.

Let's start with equations of the first type - they are the easiest to solve. And in solving them, we will be helped by such a technique as highlighting stable expressions.

Isolating a stable expression

Let's look at this equation again:

\[((4)^(x))+((4)^(x-1))=((4)^(x+1))-11\]

What do we see? The four are raised to different degrees. But all these powers are simple sums of the variable $x$ with other numbers. Therefore, it is necessary to remember the rules for working with degrees:

\[\begin(align)& ((a)^(x+y))=((a)^(x))\cdot ((a)^(y)); \\& ((a)^(x-y))=((a)^(x)):((a)^(y))=\frac(((a)^(x)))(((a )^(y))). \\\end(align)\]

Simply put, addition can be converted to a product of powers, and subtraction can easily be converted to division. Let's try to apply these formulas to the degrees from our equation:

\[\begin(align)& ((4)^(x-1))=\frac(((4)^(x)))(((4)^(1)))=((4)^ (x))\cdot \frac(1)(4); \\& ((4)^(x+1))=((4)^(x))\cdot ((4)^(1))=((4)^(x))\cdot 4. \ \\end(align)\]

Let's rewrite the original equation taking this fact into account, and then collect all the terms on the left:

\[\begin(align)& ((4)^(x))+((4)^(x))\cdot \frac(1)(4)=((4)^(x))\cdot 4 -11; \\& ((4)^(x))+((4)^(x))\cdot \frac(1)(4)-((4)^(x))\cdot 4+11=0. \\\end(align)\]

The first four terms contain the element $((4)^(x))$ - let’s take it out of the bracket:

\[\begin(align)& ((4)^(x))\cdot \left(1+\frac(1)(4)-4 \right)+11=0; \\& ((4)^(x))\cdot \frac(4+1-16)(4)+11=0; \\& ((4)^(x))\cdot \left(-\frac(11)(4) \right)=-11. \\\end(align)\]

It remains to divide both sides of the equation by the fraction $-\frac(11)(4)$, i.e. essentially multiply by the inverted fraction - $-\frac(4)(11)$. We get:

\[\begin(align)& ((4)^(x))\cdot \left(-\frac(11)(4) \right)\cdot \left(-\frac(4)(11) \right )=-11\cdot \left(-\frac(4)(11) \right); \\& ((4)^(x))=4; \\& ((4)^(x))=((4)^(1)); \\& x=1. \\\end(align)\]

That's it! We have reduced the original equation to its simplest form and obtained the final answer.

At the same time, in the process of solving we discovered (and even took it out of the bracket) the common factor $((4)^(x))$ - this is a stable expression. It can be designated as a new variable, or you can simply express it carefully and get the answer. In any case, the key principle of the solution is the following:

Find in the original equation a stable expression containing a variable that is easily distinguished from all exponential functions.

The good news is that almost every exponential equation allows you to isolate such a stable expression.

But the bad news is that these expressions can be quite tricky and can be quite difficult to identify. So let's look at one more problem:

\[((5)^(x+2))+((0,2)^(-x-1))+4\cdot ((5)^(x+1))=2\]

Perhaps someone will now have a question: “Pasha, are you stoned? There are different bases here – 5 and 0.2.” But let's try converting the power to base 0.2. For example, let’s get rid of the decimal fraction by reducing it to a regular one:

\[((0,2)^(-x-1))=((0,2)^(-\left(x+1 \right)))=((\left(\frac(2)(10 ) \right))^(-\left(x+1 \right)))=((\left(\frac(1)(5) \right))^(-\left(x+1 \right)) )\]

As you can see, the number 5 still appeared, albeit in the denominator. At the same time, the indicator was rewritten as negative. Now let’s remember one of the most important rules for working with degrees:

\[((a)^(-n))=\frac(1)(((a)^(n)))\Rightarrow ((\left(\frac(1)(5) \right))^( -\left(x+1 \right)))=((\left(\frac(5)(1) \right))^(x+1))=((5)^(x+1))\ ]

Here, of course, I was lying a little. Because for a complete understanding, the formula for getting rid of negative indicators had to be written like this:

\[((a)^(-n))=\frac(1)(((a)^(n)))=((\left(\frac(1)(a) \right))^(n ))\Rightarrow ((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(\frac(5)(1) \ right))^(x+1))=((5)^(x+1))\]

On the other hand, nothing prevented us from working with just fractions:

\[((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(((5)^(-1)) \ right))^(-\left(x+1 \right)))=((5)^(\left(-1 \right)\cdot \left(-\left(x+1 \right) \right) ))=((5)^(x+1))\]

But in this case, you need to be able to raise a power to another power (let me remind you: in this case, the indicators are added together). But I didn’t have to “reverse” the fractions - perhaps this will be easier for some. :)

In any case, the original exponential equation will be rewritten as:

\[\begin(align)& ((5)^(x+2))+((5)^(x+1))+4\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+5\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(1))\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(x+2))=2; \\& 2\cdot ((5)^(x+2))=2; \\& ((5)^(x+2))=1. \\\end(align)\]

So it turns out that the original equation can be solved even more simply than the one previously considered: here you don’t even need to select a stable expression - everything has been reduced by itself. It remains only to remember that $1=((5)^(0))$, from which we get:

\[\begin(align)& ((5)^(x+2))=((5)^(0)); \\& x+2=0; \\& x=-2. \\\end(align)\]

That's the solution! We got the final answer: $x=-2$. At the same time, I would like to note one technique that greatly simplified all calculations for us:

In exponential equations, be sure to get rid of decimal fractions and convert them to ordinary ones. This will allow you to see the same bases of degrees and greatly simplify the solution.

Let us now move on to more complex equations in which there are different bases that cannot be reduced to each other using powers at all.

Using the Degrees Property

Let me remind you that we have two more particularly harsh equations:

\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2.7)^(1-x))=0.09. \\\end(align)\]

The main difficulty here is that it is not clear what to give and to what basis. Where set expressions? Where are the same grounds? There is none of this.

But let's try to go a different way. If there are no ready-made identical bases, you can try to find them by factoring the existing bases.

Let's start with the first equation:

\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& 21=7\cdot 3\Rightarrow ((21)^(3x))=((\left(7\cdot 3 \right))^(3x))=((7)^(3x))\ cdot((3)^(3x)). \\\end(align)\]

But you can do the opposite - make the number 21 from the numbers 7 and 3. This is especially easy to do on the left, since the indicators of both degrees are the same:

\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((\left(7\cdot 3 \right))^(x+ 6))=((21)^(x+6)); \\& ((21)^(x+6))=((21)^(3x)); \\& x+6=3x; \\& 2x=6; \\& x=3. \\\end(align)\]

That's it! You took the exponent outside the product and immediately got a beautiful equation that can be solved in a couple of lines.

Now let's look at the second equation. Everything is much more complicated here:

\[((100)^(x-1))\cdot ((2.7)^(1-x))=0.09\]

\[((100)^(x-1))\cdot ((\left(\frac(27)(10) \right))^(1-x))=\frac(9)(100)\]

In this case, the fractions turned out to be irreducible, but if something could be reduced, be sure to reduce it. Often, interesting reasons will appear with which you can already work.

Unfortunately, nothing special appeared for us. But we see that the exponents on the left in the product are opposite:

Let me remind you: to get rid of the minus sign in the indicator, you just need to “flip” the fraction. Well, let's rewrite the original equation:

\[\begin(align)& ((100)^(x-1))\cdot ((\left(\frac(10)(27) \right))^(x-1))=\frac(9 )(100); \\& ((\left(100\cdot \frac(10)(27) \right))^(x-1))=\frac(9)(100); \\& ((\left(\frac(1000)(27) \right))^(x-1))=\frac(9)(100). \\\end(align)\]

In the second line, we simply took the total exponent out of the product from the bracket according to the rule $((a)^(x))\cdot ((b)^(x))=((\left(a\cdot b \right))^ (x))$, and in the last one they simply multiplied the number 100 by a fraction.

Now note that the numbers on the left (at the base) and on the right are somewhat similar. How? Yes, it’s obvious: they are powers of the same number! We have:

\[\begin(align)& \frac(1000)(27)=\frac(((10)^(3)))(((3)^(3)))=((\left(\frac( 10)(3) \right))^(3)); \\& \frac(9)(100)=\frac(((3)^(2)))(((10)^(3)))=((\left(\frac(3)(10) \right))^(2)). \\\end(align)\]

Thus, our equation will be rewritten as follows:

\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(3 )(10)\right))^(2))\]

\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(10 )(3) \right))^(3\left(x-1 \right)))=((\left(\frac(10)(3) \right))^(3x-3))\]

In this case, on the right you can also get a degree with the same base, for which it is enough to simply “turn over” the fraction:

\[((\left(\frac(3)(10) \right))^(2))=((\left(\frac(10)(3) \right))^(-2))\]

Our equation will finally take the form:

\[\begin(align)& ((\left(\frac(10)(3) \right))^(3x-3))=((\left(\frac(10)(3) \right)) ^(-2)); \\& 3x-3=-2; \\& 3x=1; \\& x=\frac(1)(3). \\\end(align)\]

That's the solution. His main idea boils down to the fact that even with different bases we try, by hook or by crook, to reduce these bases to the same thing. Elementary transformations of equations and rules for working with powers help us with this.

But what rules and when to use? How do you understand that in one equation you need to divide both sides by something, and in another you need to factor the base of the exponential function?

The answer to this question will come with experience. Try your hand at simple equations first, and then gradually complicate the problems - and very soon your skills will be enough to solve any exponential equation from the same Unified State Exam or any independent/test work.

And to help you in this difficult matter, I suggest downloading a set of equations for independent decision. All equations have answers, so you can always test yourself.

In general, I wish you a successful training. And see you in the next lesson - there we will analyze really complex exponential equations, where the methods described above are no longer enough. And simple training will not be enough either. :)

Majority decision mathematical problems is somehow related to the transformation of numerical, algebraic or functional expressions. The above applies especially to the decision. In the versions of the Unified State Exam in mathematics, this type of problem includes, in particular, task C3. Learning to solve C3 tasks is important not only for the purpose of successful passing the Unified State Exam, but also for the reason that this skill will be useful when studying a mathematics course in high school.

When completing tasks C3, you have to decide various types equations and inequalities. Among them are rational, irrational, exponential, logarithmic, trigonometric, containing modules (absolute values), as well as combined ones. This article discusses the main types of exponential equations and inequalities, as well as various methods their decisions. Read about solving other types of equations and inequalities in the “” section in articles devoted to methods for solving C3 problems from the Unified State Examination in mathematics.

Before we begin to analyze specific exponential equations and inequalities, as a math tutor, I suggest you brush up on some theoretical material that we will need.

Exponential function

What is an exponential function?

Function of the form y = a x, Where a> 0 and a≠ 1 is called exponential function.

Basic properties of exponential function y = a x:

Graph of an Exponential Function

The graph of the exponential function is exponent:

Graphs of exponential functions (exponents)

Solving exponential equations

Indicative are called equations in which the unknown variable is found only in exponents of some powers.

To solve exponential equations you need to know and be able to use the following simple theorem:

Theorem 1. Exponential equation a f(x) = a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).

In addition, it is useful to remember the basic formulas and operations with degrees:

Title="Rendered by QuickLaTeX.com">!}

Example 1. Solve the equation:

Solution: We use the above formulas and substitution:

The equation then becomes:

Discriminant of the received quadratic equation positive:

Title="Rendered by QuickLaTeX.com">!}

This means that this equation has two roots. We find them:

Moving on to reverse substitution, we get:

The second equation has no roots, since the exponential function is strictly positive over the entire domain of definition. Let's solve the second one:

Taking into account what was said in Theorem 1, we move on to the equivalent equation: x= 3. This will be the answer to the task.

Answer: x = 3.

Example 2. Solve the equation:

Solution: The equation has no restrictions on the range of permissible values, since the radical expression makes sense for any value x(exponential function y = 9 4 -x positive and not equal to zero).

We solve the equation by equivalent transformations using the rules of multiplication and division of powers:

The last transition was carried out in accordance with Theorem 1.

Answer:x= 6.

Example 3. Solve the equation:

Solution: both sides of the original equation can be divided by 0.2 x. This transition will be equivalent, since this expression is greater than zero for any value x(the exponential function is strictly positive in its domain of definition). Then the equation takes the form:

Answer: x = 0.

Example 4. Solve the equation:

Solution: we simplify the equation to an elementary one by means of equivalent transformations using the rules of division and multiplication of powers given at the beginning of the article:

Dividing both sides of the equation by 4 x, as in the previous example, is an equivalent transformation, since this expression is not equal to zero for any values x.

Answer: x = 0.

Example 5. Solve the equation:

Solution: function y = 3x, standing on the left side of the equation, is increasing. Function y = —x The -2/3 on the right side of the equation is decreasing. This means that if the graphs of these functions intersect, then at most one point. In this case, it is easy to guess that the graphs intersect at the point x= -1. There will be no other roots.

Answer: x = -1.

Example 6. Solve the equation:

Solution: we simplify the equation by means of equivalent transformations, keeping in mind everywhere that the exponential function is strictly greater than zero for any value x and using the rules for calculating the product and quotient of powers given at the beginning of the article:

Answer: x = 2.

Solving exponential inequalities

Indicative are called inequalities in which the unknown variable is contained only in exponents of some powers.

To solve exponential inequalities knowledge of the following theorem is required:

Theorem 2. If a> 1, then the inequality a f(x) > a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x). If 0< a < 1, то exponential inequality a f(x) > a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).

Example 7. Solve the inequality:

Solution: Let's present the original inequality in the form:

Let's divide both sides of this inequality by 3 2 x, in this case (due to the positivity of the function y= 3 2x) the inequality sign will not change:

Let's use the substitution:

Then the inequality will take the form:

So, the solution to the inequality is the interval:

moving to the reverse substitution, we get:

Due to the positivity of the exponential function, the left inequality is satisfied automatically. Using the well-known property of the logarithm, we move on to the equivalent inequality:

Since the base of the degree is a number greater than one, equivalent (by Theorem 2) is the transition to the following inequality:

So, we finally get answer:

Example 8. Solve the inequality:

Solution: Using the properties of multiplication and division of powers, we rewrite the inequality in the form:

Let's introduce a new variable:

Taking this substitution into account, the inequality takes the form:

Multiplying the numerator and denominator of the fraction by 7, we obtain the following equivalent inequality:

So, the following values ​​of the variable satisfy the inequality t:

Then, moving to the reverse substitution, we get:

Since the base of the degree here is greater than one, the transition to the inequality will be equivalent (by Theorem 2):

Finally we get answer:

Example 9. Solve the inequality:

Solution:

We divide both sides of the inequality by the expression:

It is always greater than zero (due to the positivity of the exponential function), so there is no need to change the inequality sign. We get:

t located in the interval:

Moving on to the reverse substitution, we find that the original inequality splits into two cases:

The first inequality has no solutions due to the positivity of the exponential function. Let's solve the second one:

Example 10. Solve the inequality:

Solution:

Parabola branches y = 2x+2-x 2 are directed downwards, therefore it is limited from above by the value that it reaches at its vertex:

Parabola branches y = x 2 -2x The +2 in the indicator are directed upward, which means it is limited from below by the value that it reaches at its vertex:

At the same time, the function also turns out to be bounded from below y = 3 x 2 -2x+2, which is on the right side of the equation. She achieves her goal lowest value at the same point as the parabola in the exponent, and this value is equal to 3 1 = 3. So, the original inequality can only be true if the function on the left and the function on the right take on a value equal to 3 at the same point (by the intersection The range of values ​​of these functions is only this number). This condition is satisfied at a single point x = 1.

Answer: x= 1.

In order to learn to decide exponential equations and inequalities, it is necessary to constantly train in solving them. Various things can help you with this difficult task. methodological manuals, problem books on elementary mathematics, collections of competitive problems, mathematics classes at school, as well as individual lessons with a professional tutor. I sincerely wish you success in your preparation and excellent results in the exam.

Sergey Valerievich

P.S. Dear guests! Please do not write requests to solve your equations in the comments. Unfortunately, I have absolutely no time for this. Such messages will be deleted. Please read the article. Perhaps in it you will find answers to questions that did not allow you to solve your task on your own.