Preparation for the Unified State Exam. Exponential equations

Go to the youtube channel of our website to stay up to date with all the new video lessons.

First, let's remember the basic formulas of powers and their properties.

Product of a number a occurs on itself n times, we can write this expression as a a … a=a n

1. a 0 = 1 (a ≠ 0)

3. a n a m = a n + m

4. (a n) m = a nm

5. a n b n = (ab) n

7. a n / a m = a n - m

Power or exponential equations– these are equations in which the variables are in powers (or exponents), and the base is a number.

Examples of exponential equations:

In this example, the number 6 is the base; it is always at the bottom, and the variable x degree or indicator.

Let us give more examples of exponential equations.
2 x *5=10
16 x - 4 x - 6=0

Now let's look at how exponential equations are solved?

Let's take a simple equation:

2 x = 2 3

This example can be solved even in your head. It can be seen that x=3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let’s see how to formalize this decision:

2 x = 2 3
x = 3

In order to solve such an equation, we removed identical grounds(that is, twos) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our decision.

Algorithm for solving the exponential equation:
1. Need to check identical whether the equation has bases on the right and left. If the reasons are not the same, we are looking for options to solve this example.
2. After the bases become the same, equate degrees and solve the resulting new equation.

Now let's look at a few examples:

Let's start with something simple.

The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their degrees.

x+2=4 The simplest equation is obtained.
x=4 – 2
x=2
Answer: x=2

In the following example you can see that the bases are different: 3 and 9.

3 3x - 9 x+8 = 0

First, move the nine to the right side, we get:

Now you need to make the same bases. We know that 9=3 2. Let's use the power formula (a n) m = a nm.

3 3x = (3 2) x+8

We get 9 x+8 =(3 2) x+8 =3 2x+16

3 3x = 3 2x+16 Now it is clear that on the left and right sides the bases are the same and equal to three, which means we can discard them and equate the degrees.

3x=2x+16 we get the simplest equation
3x - 2x=16
x=16
Answer: x=16.

Let's look at the following example:

2 2x+4 - 10 4 x = 2 4

First of all, we look at the bases, bases two and four. And we need them to be the same. We transform the four using the formula (a n) m = a nm.

4 x = (2 2) x = 2 2x

And we also use one formula a n a m = a n + m:

2 2x+4 = 2 2x 2 4

Add to the equation:

2 2x 2 4 - 10 2 2x = 24

We gave an example for the same reasons. But other numbers 10 and 24 bother us. What to do with them? If you look closely you can see that on the left side we have 2 2x repeated, here is the answer - we can put 2 2x out of brackets:

2 2x (2 4 - 10) = 24

Let's calculate the expression in brackets:

2 4 — 10 = 16 — 10 = 6

We divide the entire equation by 6:

Let's imagine 4=2 2:

2 2x = 2 2 bases are the same, we discard them and equate the degrees.
2x = 2 is the simplest equation. Divide it by 2 and we get
x = 1
Answer: x = 1.

Let's solve the equation:

9 x – 12*3 x +27= 0

Let's convert:
9 x = (3 2) x = 3 2x

We get the equation:
3 2x - 12 3 x +27 = 0

Our bases are the same, equal to three. In this example, you can see that the first three has a degree twice (2x) than the second (just x). In this case, you can solve replacement method. We replace the number with the smallest degree:

Then 3 2x = (3 x) 2 = t 2

We replace all x powers in the equation with t:

t 2 - 12t+27 = 0
We get a quadratic equation. Solving through the discriminant, we get:
D=144-108=36
t 1 = 9
t2 = 3

Returning to the variable x.

Take t 1:
t 1 = 9 = 3 x

Therefore,

3 x = 9
3 x = 3 2
x 1 = 2

One root was found. We are looking for the second one from t 2:
t 2 = 3 = 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 = 2; x 2 = 1.

On the website you can ask any questions you may have in the HELP DECIDE section, we will definitely answer you.

Join the group

At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “ Exponential equations" The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of task, graduates will be able to count on high scores when passing the Unified State Examination in mathematics.

Get ready for exam testing with Shkolkovo!

When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. School textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.

The Shkolkovo educational portal invites students to use our knowledge base. We implement completely new method preparation for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.

Shkolkovo teachers have collected, systematized and presented everything necessary for successful passing the Unified State Exam material in the simplest and most accessible form.

Basic definitions and formulas are presented in the “Theoretical background” section.

To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples of exponential equations with solutions presented on this page to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or move straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.

Those examples with indicators that caused you difficulties can be added to “Favorites”. This way you can quickly find them and discuss the solution with your teacher.

To successfully pass the Unified State Exam, study on the Shkolkovo portal every day!

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 do you need to 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 I could: 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 exponentials 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 “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 exponential function, to which we 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 - they are just different forms of writing 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:

Write down the original equation. For example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
Do some weird shit. Or even some crap called "convert an equation";
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:

The equation is composed of exponential functions with the same base. Example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
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 now has 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 decimals, convert them to regular ones. This will allow you to see the same bases of degrees and greatly simplify the solution.

Let's move on now to more complex equations, in which there are different bases that are not at all reducible to each other using degrees.

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. :)

Back Forward

Attention! Slide previews are for informational purposes only and may not represent all of the presentation's features. If you are interested in this work, please download the full version.

Lesson type

: lesson on generalization and complex application of knowledge, skills and abilities on the topic “Exponential equations and methods for solving them.”

Lesson objectives.

Educational:
repeat and systematize the main material of the topic “Exponential equations, their solutions”; consolidate the ability to use appropriate algorithms when solving exponential equations of various types; preparation for the Unified State Exam.
Educational:
develop students’ logical and associative thinking; promote the development of the skill of independent application of knowledge.
Educational:
cultivate dedication, attention and accuracy when solving equations.

Equipment:

computer and multimedia projector.

Used in class information Technology : methodological support to the lesson - presentation in Microsoft Power Point.

Lesson progress

Every skill comes with hard work

I. Setting a lesson goal(Slide number 2 )

In this lesson, we will summarize and generalize the topic “Exponential equations, their solutions.” Let's get acquainted with typical Unified State Exam assignments different years on this topic.

Problems on solving exponential equations can be found in any part of the Unified State Examination tasks. In the part “ IN " Usually they offer to solve the simplest exponential equations. In the part “ WITH " You can find more complex exponential equations, the solution of which is usually one of the stages of completing the task.

For example ( Slide number 3 ).

Unified State Examination - 2007

Q 4 – Find the greatest value of the expression x y, Where ( X; at) – solution of the system:

Unified State Examination - 2008

Q 1 – Solve the equations:

A) X 6 3X – 36 6 3X = 0;

b) 4 X +1 + 8 4X= 3.

Unified State Examination - 2009

Q 4 – Find the meaning of the expression x + y, Where ( X; at) – solution of the system:

Unified State Examination - 2010

– Solve the equation: 7 X– 2 = 49. – Find the roots of the equation: 4 X 2 + 3X – 2 - 0,5 2x2 + 2X – 1 = 0. – Solve the system of equations:

II. Updating basic knowledge. Repetition

(Slides No. 4 – 6 presentations for the lesson)

Shown on screen background summary of theoretical material on the topic.

The following issues are discussed:

What equations are called indicative?
Name the main ways to solve them. Give examples of their types ( Slide number 4 )

(Independently solve the proposed equations for each method and perform a self-test using the slide)

What theorem is used when solving simple exponential equations of the form: and f(x) = a g(x) ?
What other methods for solving exponential equations exist? ( Slide number 5 )

Factorization method

Method of division (multiplication) by an exponential expression other than zero when solving homogeneous exponential equations

Advice:

When solving exponential equations, it is useful to first make transformations, obtaining powers with the same bases on both sides of the equation.

Solving equations using the last two methods with subsequent comments

(Slide number 6 ).

. 4 X+ 1 – 2 4 X– 2 = 124, 4 X– 2 (4 3 - 2) = 124, 4 X– 2 62 = 124,

4 X– 2 = 2, 4 X– 2 = 4 0,5 , X– 2 = 0,5, x = 2,5 .

2 2 2х – 3 2 X 5X - 5 5 2X= 0¦: 5 2 X 0,

2 (2/5) 2х – 3 (2/5) X - 5 = 0,

t = (2/5) x, t > 0, 2t 2 - 3t- 5 = 0,t= -1(?...), t = 5/2; 5/2 = (2/5) x, X= ?...

III. Solving Unified State Exam 2010 tasks

Students independently solve the tasks proposed at the beginning of the lesson on slide No. 3, using instructions for the solution, check their progress in solving and answers to them using a presentation ( Slide number 7). During the work, options and solutions are discussed, and attention is drawn to possible errors in the solution.

: a) 7 X– 2 = 49, b) (1/6) 12 – 7 x = 36. Answer: A) X= 4, b) X = 2. : 4 X 2 + 3X – 2 - 0,5 2x2 + 2X– 1 = 0. (Can be replaced by 0.5 = 4 – 0.5)

Solution. ,

X 2 + 3X – 2 = -X 2 - 4X + 0,5 …

Answer: X= -5/2, X = 1/2.

: 5 5 tg y+ 4 = 5 -tg y, at cos y< 0.

Directions to the solution

. 5 5 tg y+ 4 = 5 -tg y¦ 5 tg y 0,

5 5 2g y+ 4 5 tg y – 1 = 0. Let X= 5 tg y , …

5 tg y = -1 (?...), 5 tg y = 1/5.

Since tg y= -1 and cos y< 0, then at II coordinate quarter

Answer: at= 3/4 + 2k, k N.

IV. Teamwork at the board

A high level of training task is being considered - Slide number 8. With the help of this slide, a dialogue between the teacher and students occurs, facilitating the development of a solution.

– At what parameter A equation 2 2 X – 3 2 X + A 2 – 4A= 0 has two roots?

Let t= 2 X, Where t > 0 . We get t 2 – 3t + (A 2 – 4A) = 0 .

1). Since the equation has two roots, then D > 0;

2). Because t 1,2 > 0, then t 1 t 2 > 0, that is A 2 – 4A> 0 (?...).

Answer: A(– 0.5; 0) or (4; 4.5).

V. Test work

(Slide number 9 )

Students perform test work on pieces of paper, exercising self-monitoring and self-evaluation of the work performed using a presentation, becoming established in the topic. They independently determine for themselves a program for regulating and correcting knowledge based on mistakes made in workbooks. Sheets with completed independent work are handed over to the teacher for checking.

The underlined numbers are of a basic level, those with an asterisk are of increased complexity.