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Exponential equations. Comprehensive Guide (2019)

Hello! Today we will discuss with you how to solve equations that can be either elementary (and I hope that after reading this article, almost all of them will be so for you), and those that are usually given “for filling”. Apparently to finally fall asleep. But I will try to do everything possible so that now you don’t get into trouble when faced with this type of equations. I won't beat around the bush anymore, I'll open it right away little secret: today we will study exponential equations.

Before moving on to analyzing ways to solve them, I will immediately outline for you a range of questions (quite small) that you should repeat before rushing to attack this topic. So, 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. Which one? Right: . 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, we did this in the previous example: 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 less, the less 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, everything leads to the smallest base: , . Then the original equation will be equivalent to the following: All I need is to use the properties of powers: 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 for different reasons, but with the same indicators:

The left side of the equation will look like: 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 that you can skip a degree with a negative exponent? Well, here I am about the same thing as 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)

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 these ones:

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 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. Now I want to give you a few life examples, which will help you understand why they are needed in principle. Here I will give two examples. One of them is quite everyday, but the other is more likely to be of scientific rather than practical interest.

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 be the initial sum, - final amount, - interest rate for the period, - number of periods. Then:

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, in order to receive a million, we will need to make a deposit for a month (not very quickly, right?).

Example 2 (rather scientific). Despite his certain “isolation”, I recommend that you pay attention to him: he regularly “slips into the Unified State Examination!! (the problem is taken from the “real” version) During the decay of a 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 completely real application on practice. Now I want to show you another (simple) way to solve exponential equations, which is 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 to factor the expression:

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 brackets, and then - come what may, I believe that we will be lucky =)) For example:

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

Here's another example for you to reinforce:

I will give his brief solution (without bothering myself much 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. I will just give brief recommendations and tips for solving them:

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. AVERAGE 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 simple examples.

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

“method of 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 the 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:

This equation is solved using a “simple substitution,” 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.

It is clear that most likely we will have to make a replacement (this is the smallest of the powers included in our equation), but 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: a cubic equation with absolutely terrible formulas for solving it (well, speaking in general terms). 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 part: !
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? That'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? By itself: .

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 did not at all want to scare you with this example; rather, my goal was to show that although we had a fairly simple replacement, it nevertheless led to quite a complex equation, the solution of which required some special skills from us. Well, no one is immune from this. But the replacement in in this case was pretty obvious.

Here's an example with a slightly 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. The following tasks are taken from the Unified State Examination C1 ( increased level difficulties). 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:

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.
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.
3. The third equation is also solved quite standardly: 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:

   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 see, selection of roots of exponential equations requires sufficient deep knowledge 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. Let's practice with one more example:

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 the first root. Let's compare and: since, then. (property 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:

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. Where it leads? 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 note 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 are resolved " 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.

EXPONENTARY EQUATIONS. 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.

Exponential equations. As you know, the Unified State Examination includes simple equations. We have already considered some - these are logarithmic, trigonometric, rational. Here are the exponential equations.

In a recent article we worked with exponential expressions, it will be useful. The equations themselves are solved simply and quickly. You just need to know the properties of exponents and... About thisFurther.

Let us list the properties of exponents:

The zero power of any number is equal to one.

A corollary from this property:

A little more theory.

An exponential equation is an equation containing a variable in the exponent, that is, it is an equation of the form:

f(x) expression that contains a variable

Methods for solving exponential equations

1. As a result of transformations, the equation can be reduced to the form:

Then we apply the property:

2. Upon obtaining an equation of the form a f (x) = b using the definition of logarithm, we get:

3. As a result of transformations, you can obtain an equation of the form:

Logarithm applied:

Express and find x.

In tasks Unified State Exam options It will be enough to use the first method.

That is, it is necessary to represent the left and right sides in the form of powers with the same basis, and then we equate the exponents and solve the usual linear equation.

Consider the equations:

Find the root of equation 4 1–2x = 64.

It is necessary to ensure that the left and right sides contain exponential expressions with the same base. We can represent 64 as 4 to the power of 3. We get:

4 1–2x = 4 3

1 – 2x = 3

– 2x = 2

x = – 1

Examination:

4 1–2 (–1) = 64

4 1 + 2 = 64

4 3 = 64

64 = 64

Answer: –1

Find the root of equation 3 x–18 = 1/9.

It is known that

So 3 x-18 = 3 -2

The bases are equal, we can equate the indicators:

x – 18 = – 2

x = 16

Examination:

3 16–18 = 1/9

3 –2 = 1/9

1/9 = 1/9

Answer: 16

Find the root of the equation:

Let's represent the fraction 1/64 as one-fourth to the third power:

2x – 19 = 3

2x = 22

x = 11

Examination:

Answer: 11

Find the root of the equation:

Let's imagine 1/3 as 3 –1, and 9 as 3 squared, we get:

(3 –1) 8–2x = 3 2

3 –1∙(8–2x) = 3 2

3 –8+2x = 3 2

Now we can equate the indicators:

– 8+2x = 2

2x = 10

x = 5

Examination:

Answer: 5

26654. Find the root of the equation:

Solution:

Answer: 8.75

Indeed, no matter what power we raise a positive number a to, we cannot get a negative number.

Any exponential equation after appropriate transformations is reduced to solving one or more simple ones.In this section we will also look at solving some equations, don’t miss it!That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

Examples:

\(4^x=32\)
\(5^(2x-1)-5^(2x-3)=4.8\)
\((\sqrt(7))^(2x+2)-50\cdot(\sqrt(7))^(x)+7=0\)

How to Solve Exponential Equations

When solving any exponential equation, we strive to bring it to the form \(a^(f(x))=a^(g(x))\), and then make the transition to equality of exponents, that is:

\(a^(f(x))=a^(g(x))\) \(⇔\) \(f(x)=g(x)\)

For example:\(2^(x+1)=2^2\) \(⇔\) \(x+1=2\)

Important! From the same logic, two requirements for such a transition follow:
- number in left and right should be the same;
- the degrees on the left and right must be “pure”, that is, there should be no multiplications, divisions, etc.

For example:

To reduce the equation to the form \(a^(f(x))=a^(g(x))\) and are used.

Example . Solve the exponential equation \(\sqrt(27)·3^(x-1)=((\frac(1)(3)))^(2x)\)
Solution:

\(\sqrt(27)·3^(x-1)=((\frac(1)(3)))^(2x)\)		We know that \(27 = 3^3\). Taking this into account, let us transform the equation.
\(\sqrt(3^3)·3^(x-1)=((\frac(1)(3)))^(2x)\)		By the property of the root \(\sqrt[n](a)=a^(\frac(1)(n))\) we obtain that \(\sqrt(3^3)=((3^3))^( \frac(1)(2))\). Next, using the property of degree \((a^b)^c=a^(bc)\), we obtain \(((3^3))^(\frac(1)(2))=3^(3 \ cdot \frac(1)(2))=3^(\frac(3)(2))\).
\(3^(\frac(3)(2))\cdot 3^(x-1)=(\frac(1)(3))^(2x)\)		We also know that \(a^b·a^c=a^(b+c)\). Applying this to the left side, we get: \(3^(\frac(3)(2))·3^(x-1)=3^(\frac(3)(2)+ x-1)=3^ (1.5 + x-1)=3^(x+0.5)\).
\(3^(x+0.5)=(\frac(1)(3))^(2x)\)		Now remember that: \(a^(-n)=\frac(1)(a^n)\). This formula can also be used in reverse side: \(\frac(1)(a^n) =a^(-n)\). Then \(\frac(1)(3)=\frac(1)(3^1) =3^(-1)\).
\(3^(x+0.5)=(3^(-1))^(2x)\)		Applying the property \((a^b)^c=a^(bc)\) to the right side, we obtain: \((3^(-1))^(2x)=3^((-1) 2x) =3^(-2x)\).
\(3^(x+0.5)=3^(-2x)\)		And now our bases are equal and there are no interfering coefficients, etc. So we can make the transition.
		Example . Solve the exponential equation \(4^(x+0.5)-5 2^x+2=0\) Solution: \(4^(x+0.5)-5 2^x+2=0\) We again use the power property \(a^b \cdot a^c=a^(b+c)\) in the opposite direction. \(4^x 4^(0.5)-5 2^x+2=0\) Now remember that \(4=2^2\). \((2^2)^x·(2^2)^(0.5)-5·2^x+2=0\) Using the properties of degrees, we transform: \((2^2)^x=2^(2x)=2^(x 2)=(2^x)^2\) \((2^2)^(0.5)=2^(2 0.5)=2^1=2.\) \(2·(2^x)^2-5·2^x+2=0\) We look carefully at the equation and see that the replacement \(t=2^x\) suggests itself. \(t_1=2\) \(t_2=\frac(1)(2)\) However, we found the values ​​of \(t\), and we need \(x\). We return to the X's, making a reverse replacement. \(2^x=2\) \(2^x=\frac(1)(2)\) Let's transform the second equation using the property of negative power... \(2^x=2^1\) \(2^x=2^(-1)\) ...and we decide until the answer. \(x_1=1\) \(x_2=-1\) Answer : \(-1; 1\). The question remains - how to understand when to use which method? This comes with experience. Until you get it, use it general recommendation for solutions complex tasks- “If you don’t know what to do, do what you can.” That is, look for how you can transform the equation in principle, and try to do it - what if what happens? The main thing is to make only mathematically based transformations. Exponential equations without solutions Let's look at two more situations that often confuse students: - a positive number to the power is equal to zero, for example, \(2^x=0\); - a positive number to the power is equal to negative number, for example, \(2^x=-4\). Let's try to solve by brute force. If x is a positive number, then as x grows, the entire power \(2^x\) will only increase: \(x=1\); \(2^1=2\) \(x=2\); \(2^2=4\) \(x=3\); \(2^3=8\). \(x=0\); \(2^0=1\) Also by. Negative X's remain. Recalling the property \(a^(-n)=\frac(1)(a^n)\), we check: \(x=-1\); \(2^(-1)=\frac(1)(2^1) =\frac(1)(2)\) \(x=-2\); \(2^(-2)=\frac(1)(2^2) =\frac(1)(4)\) \(x=-3\); \(2^(-3)=\frac(1)(2^3) =\frac(1)(8)\) Despite the fact that the number becomes smaller with each step, it will never reach zero. So the negative degree did not save us. We come to a logical conclusion: A positive number to any degree will remain a positive number. Thus, both equations above have no solutions. Exponential equations with different bases In practice, sometimes we encounter exponential equations with different bases that are not reducible to each other, and at the same time with the same exponents. They look like this: \(a^(f(x))=b^(f(x))\), where \(a\) and \(b\) are positive numbers. For example: \(7^(x)=11^(x)\) \(5^(x+2)=3^(x+2)\) \(15^(2x-1)=(\frac(1)(7))^(2x-1)\) Such equations can easily be solved by dividing by any of the sides of the equation (usually divided by the right side, that is, by \(b^(f(x))\). You can divide this way because a positive number is positive to any power (that is, we do not divide by zero). \(\frac(a^(f(x)))(b^(f(x)))\) \(=1\) Example . Solve the exponential equation \(5^(x+7)=3^(x+7)\) Solution: \(5^(x+7)=3^(x+7)\) Here we won’t be able to turn a five into a three, or vice versa (at least without using ). This means we cannot come to the form \(a^(f(x))=a^(g(x))\). However, the indicators are the same. Let's divide the equation by the right side, that is, by \(3^(x+7)\) (we can do this because we know that three will not be zero to any degree). \(\frac(5^(x+7))(3^(x+7))\) \(=\)\(\frac(3^(x+7))(3^(x+7) )\) Now remember the property \((\frac(a)(b))^c=\frac(a^c)(b^c)\) and use it from the left in the opposite direction. On the right, we simply reduce the fraction. \((\frac(5)(3))^(x+7)\) \(=1\) It would seem that things didn't get any better. But remember one more property of power: \(a^0=1\), in other words: “any number to the zero power is equal to \(1\).” The converse is also true: “one can be represented as any number to the zero power.” Let's take advantage of this by making the base on the right the same as on the left. \((\frac(5)(3))^(x+7)\) \(=\) \((\frac(5)(3))^0\) Voila! Let's get rid of the bases. We are writing a response. Answer : \(-7\). Sometimes the “sameness” of exponents is not obvious, but skillful use of the properties of exponents resolves this issue. Example . Solve the exponential equation \(7^( 2x-4)=(\frac(1)(3))^(-x+2)\) Solution: \(7^( 2x-4)=(\frac(1)(3))^(-x+2)\) The equation looks very sad... Not only can the grounds not be reduced to the same number(seven will in no way be equal to \(\frac(1)(3)\)), so the exponents are also different... However, let’s use two in the exponent of the left power. \(7^( 2(x-2))=(\frac(1)(3))^(-x+2)\) Remembering the property \((a^b)^c=a^(b·c)\) , we transform from the left: \(7^(2(x-2))=7^(2·(x-2))=(7^2)^(x-2)=49^(x-2)\). \(49^(x-2)=(\frac(1)(3))^(-x+2)\) Now, remembering the property of negative degree \(a^(-n)=\frac(1)(a)^n\), we transform from the right: \((\frac(1)(3))^(-x+2) =(3^(-1))^(-x+2)=3^(-1(-x+2))=3^(x-2)\) \(49^(x-2)=3^(x-2)\) Hallelujah! The indicators are the same! Acting according to the scheme already familiar to us, we solve before the answer. Answer : \(2\).

Lecture: “Methods for solving exponential equations.”

1 . Exponential equations.

Equations containing unknowns in exponents are called exponential equations. The simplest of them is the equation ax = b, where a > 0, a ≠ 1.

1) At b< 0 и b = 0 это уравнение, согласно свойству 1 exponential function, has no solution.

2) For b > 0, using the monotonicity of the function and the root theorem, the equation has a unique root. In order to find it, b must be represented in the form b = aс, аx = bс ó x = c or x = logab.

Exponential equations by algebraic transformations lead to standard equations, which are solved using the following methods:

1) method of reduction to one base;

2) assessment method;

3) graphic method;

4) method of introducing new variables;

5) factorization method;

6) exponential – power equations;

7) demonstrative with a parameter.

2 . Method of reduction to one base.

The method is based on following property degrees: if two degrees are equal and their bases are equal, then their exponents are equal, that is, we must try to reduce the equation to the form

Examples. Solve the equation:

1 . 3x = 81;

Let's represent the right side of the equation in the form 81 = 34 and write the equation equivalent to the original 3 x = 34; x = 4. Answer: 4.

2. https://pandia.ru/text/80/142/images/image004_8.png" width="52" height="49">and let's move on to the equation for exponents 3x+1 = 3 – 5x; 8x = 4; x = 0.5. Answer: 0.5.

3. https://pandia.ru/text/80/142/images/image006_8.png" width="105" height="47">

Note that the numbers 0.2, 0.04, √5 and 25 represent powers of 5. Let's take advantage of this and transform the original equation as follows:

, whence 5-x-1 = 5-2x-2 ó - x – 1 = - 2x – 2, from which we find the solution x = -1. Answer: -1.

5. 3x = 5. By definition of logarithm, x = log35. Answer: log35.

6. 62x+4 = 33x. 2x+8.

Let's rewrite the equation in the form 32x+4.22x+4 = 32x.2x+8, i.e..png" width="181" height="49 src="> Hence x – 4 =0, x = 4. Answer: 4.

7 . 2∙3x+1 - 6∙3x-2 - 3x = 9. Using the properties of powers, we write the equation in the form 6∙3x - 2∙3x – 3x = 9 then 3∙3x = 9, 3x+1 = 32, i.e. i.e. x+1 = 2, x =1. Answer: 1.

Problem bank No. 1.

Solve the equation:

Test No. 1.

	1) 0 2) 4 3) -2 4) -4
A2 32x-8 = √3.	1)17/4 2) 17 3) 13/2 4) -17/4
A3	1) 3;1 2) -3;-1 3) 0;2 4) no roots
	1) 7;1 2) no roots 3) -7;1 4) -1;-7
A5	1) 0;2; 2) 0;2;3 3) 0 4) -2;-3;0
A6	1) -1 2) 0 3) 2 4) 1

Test No. 2

A1	1) 3 2) -1;3 3) -1;-3 4) 3;-1
A2	1) 14/3 2) -14/3 3) -17 4) 11
A3	1) 2;-1 2) no roots 3) 0 4) -2;1
A4	1) -4 2) 2 3) -2 4) -4;2
A5	1) 3 2) -3;1 3) -1 4) -1;3

3 Evaluation method.

Root theorem: if the function f(x) increases (decreases) on the interval I, the number a is any value taken by f on this interval, then the equation f(x) = a has a single root on the interval I.

When solving equations using the estimation method, this theorem and the monotonicity properties of the function are used.

Examples. Solve equations: 1. 4x = 5 – x.

Solution. Let's rewrite the equation as 4x +x = 5.

1. if x = 1, then 41+1 = 5, 5 = 5 is true, which means 1 is the root of the equation.

Function f(x) = 4x – increases on R, and g(x) = x – increases on R => h(x)= f(x)+g(x) increases on R, as the sum of increasing functions, then x = 1 is the only root of the equation 4x = 5 – x. Answer: 1.

2.

Solution. Let's rewrite the equation in the form .

1. if x = -1, then , 3 = 3 is true, which means x = -1 is the root of the equation.

2. prove that he is the only one.

3. Function f(x) = - decreases on R, and g(x) = - x – decreases on R=> h(x) = f(x)+g(x) – decreases on R, as the sum of decreasing functions . This means, by the root theorem, x = -1 is the only root of the equation. Answer: -1.

Problem bank No. 2. Solve the equation

a) 4x + 1 =6 – x;

b)

c) 2x – 2 =1 – x;

4. Method of introducing new variables.

The method is described in paragraph 2.1. The introduction of a new variable (substitution) is usually carried out after transformations (simplification) of the terms of the equation. Let's look at examples.

Examples. R Solve the equation: 1. .

Let's rewrite the equation differently: https://pandia.ru/text/80/142/images/image030_0.png" width="128" height="48 src="> i.e..png" width="210" height ="45">

Solution. Let's rewrite the equation differently:

Let's designate https://pandia.ru/text/80/142/images/image035_0.png" width="245" height="57"> - not suitable.

t = 4 => https://pandia.ru/text/80/142/images/image037_0.png" width="268" height="51"> - irrational equation. We note that

The solution to the equation is x = 2.5 ≤ 4, which means 2.5 is the root of the equation. Answer: 2.5.

Solution. Let's rewrite the equation in the form and divide both sides by 56x+6 ≠ 0. We get the equation

2x2-6x-7 = 2x2-6x-8 +1 = 2(x2-3x-4)+1, t..png" width="118" height="56">

The roots of the quadratic equation are t1 = 1 and t2<0, т. е..png" width="200" height="24">.

Solution . Let's rewrite the equation in the form

and note that it is a homogeneous equation of the second degree.

Divide the equation by 42x, we get

Let's replace https://pandia.ru/text/80/142/images/image049_0.png" width="16" height="41 src="> .

Answer: 0; 0.5.

Problem bank No. 3. Solve the equation

b)

G)

Test No. 3 with a choice of answers. Minimum level.

A1	1) -0.2;2 2) log52 3) –log52 4) 2
A2 0.52x – 3 0.5x +2 = 0.	1) 2;1 2) -1;0 3) no roots 4) 0
	1) 0 2) 1; -1/3 3) 1 4) 5
A4 52x-5x - 600 = 0.	1) -24;25 2) -24,5; 25,5 3) 25 4) 2
	1) no roots 2) 2;4 3) 3 4) -1;2

Test No. 4 with a choice of answers. General level.

A1	1) 2;1 2) ½;0 3)2;0 4) 0
A2 2x – (0.5)2x – (0.5)x + 1 = 0	1) -1;1 2) 0 3) -1;0;1 4) 1
	1) 64 2) -14 3) 3 4) 8
	1)-1 2) 1 3) -1;1 4) 0
A5	1) 0 2) 1 3) 0;1 4) no roots

5. Factorization method.

1. Solve the equation: 5x+1 - 5x-1 = 24.

Solution..png" width="169" height="69"> , from where

2. 6x + 6x+1 = 2x + 2x+1 + 2x+2.

Solution. Let's put 6x out of brackets on the left side of the equation, and 2x on the right side. We get the equation 6x(1+6) = 2x(1+2+4) ó 6x = 2x.

Since 2x >0 for all x, we can divide both sides of this equation by 2x without fear of losing solutions. We get 3x = 1ó x = 0.

3.

Solution. Let's solve the equation using the factorization method.

Let us select the square of the binomial

4. https://pandia.ru/text/80/142/images/image067_0.png" width="500" height="181">

x = -2 is the root of the equation.

Equation x + 1 = 0 " style="border-collapse:collapse;border:none">

A1 5x-1 +5x -5x+1 =-19.

1) 1 2) 95/4 3) 0 4) -1

A2 3x+1 +3x-1 =270.

1) 2 2) -4 3) 0 4) 4

A3 32x + 32x+1 -108 = 0. x=1.5

1) 0,2 2) 1,5 3) -1,5 4) 3

1) 1 2) -3 3) -1 4) 0

A5 2x -2x-4 = 15. x=4

1) -4 2) 4 3) -4;4 4) 2

Test No. 6 General level.

A1 (22x-1)(24x+22x+1)=7.	1) ½ 2) 2 3) -1;3 4) 0.2
A2	1) 2.5 2) 3;4 3) log43/2 4) 0
A3 2x-1-3x=3x-1-2x+2.	1) 2 2) -1 3) 3 4) -3
A4	1) 1,5 2) 3 3) 1 4) -4
A5	1) 2 2) -2 3) 5 4) 0

6. Exponential – power equations.

Adjacent to exponential equations are the so-called exponential-power equations, i.e., equations of the form (f(x))g(x) = (f(x))h(x).

If it is known that f(x)>0 and f(x) ≠ 1, then the equation, like the exponential one, is solved by equating the exponents g(x) = f(x).

If the condition does not exclude the possibility of f(x)=0 and f(x)=1, then we have to consider these cases when solving an exponential equation.

1..png" width="182" height="116 src=">

2.

Solution. x2 +2x-8 – makes sense for any x, because it is a polynomial, which means the equation is equivalent to the totality

https://pandia.ru/text/80/142/images/image078_0.png" width="137" height="35">

b)

7. Exponential equations with parameters.

1. For what values ​​of the parameter p does equation 4 (5 – 3)2 +4p2–3p = 0 (1) have a unique solution?

Solution. Let us introduce the replacement 2x = t, t > 0, then equation (1) will take the form t2 – (5p – 3)t + 4p2 – 3p = 0. (2)

Discriminant of equation (2) D = (5p – 3)2 – 4(4p2 – 3p) = 9(p – 1)2.

Equation (1) has a unique solution if equation (2) has one positive root. This is possible in the following cases.

1. If D = 0, that is, p = 1, then equation (2) will take the form t2 – 2t + 1 = 0, hence t = 1, therefore, equation (1) has a unique solution x = 0.

2. If p1, then 9(p – 1)2 > 0, then equation (2) has two different roots t1 = p, t2 = 4p – 3. The conditions of the problem are satisfied by a set of systems

Substituting t1 and t2 into the systems, we have

https://pandia.ru/text/80/142/images/image084_0.png" alt="no35_11" width="375" height="54"> в зависимости от параметра a?!}

Solution. Let then equation (3) will take the form t2 – 6t – a = 0. (4)

Let us find the values ​​of the parameter a for which at least one root of equation (4) satisfies the condition t > 0.

Let us introduce the function f(t) = t2 – 6t – a. The following cases are possible.

https://pandia.ru/text/80/142/images/image087.png" alt="http://1september.ru/ru/mat/2002/35/no35_14.gif" align="left" width="215" height="73 src=">где t0 - абсцисса вершины параболы и D - дискриминант !} quadratic trinomial f(t);

https://pandia.ru/text/80/142/images/image089.png" alt="http://1september.ru/ru/mat/2002/35/no35_16.gif" align="left" width="60" height="51 src=">!}

Case 2. Equation (4) has a unique positive solution if

D = 0, if a = – 9, then equation (4) will take the form (t – 3)2 = 0, t = 3, x = – 1.

Case 3. Equation (4) has two roots, but one of them does not satisfy the inequality t > 0. This is possible if

https://pandia.ru/text/80/142/images/image092.png" alt="no35_17" width="267" height="63">!}

Thus, for a 0, equation (4) has a single positive root . Then equation (3) has a unique solution

When a< – 9 уравнение (3) корней не имеет.

if a< – 9, то корней нет; если – 9 < a < 0, то
if a = – 9, then x = – 1;

if a  0, then

Let us compare the methods for solving equations (1) and (3). Note that when solving equation (1) was reduced to a quadratic equation, the discriminant of which is a perfect square; Thus, the roots of equation (2) were immediately calculated using the formula for the roots of a quadratic equation, and then conclusions were drawn regarding these roots. Equation (3) has been reduced to a quadratic equation (4), the discriminant of which is not a perfect square, therefore, when solving equation (3), it is advisable to use theorems on the location of the roots of a quadratic trinomial and a graphical model. Note that equation (4) can be solved using Vieta's theorem.

Let's solve more complex equations.

Problem 3: Solve the equation

Solution. ODZ: x1, x2.

Let's introduce a replacement. Let 2x = t, t > 0, then as a result of transformations the equation will take the form t2 + 2t – 13 – a = 0. (*) Let us find the values ​​of a for which at least one root of the equation (*) satisfies the condition t > 0.

https://pandia.ru/text/80/142/images/image098.png" alt="http://1september.ru/ru/mat/2002/35/no35_23.gif" align="left" width="71" height="68 src=">где t0 - абсцисса вершины f(t) = t2 + 2t – 13 – a, D - дискриминант квадратного трехчлена f(t).!}

https://pandia.ru/text/80/142/images/image100.png" alt="http://1september.ru/ru/mat/2002/35/no35_25.gif" align="left" width="360" height="32 src=">!}

https://pandia.ru/text/80/142/images/image102.png" alt="http://1september.ru/ru/mat/2002/35/no35_27.gif" align="left" width="218" height="42 src=">!}

Answer: if a > – 13, a  11, a  5, then if a – 13,

a = 11, a = 5, then there are no roots.

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Mathematics at school No. 2, 1987 pp. 9 – 11.

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entering universities. “A S T - press school”, 2002

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Minsk and Russian Federation “Review”, 1996

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M. "Intellect - Center", 2003

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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 problem, graduates can count on high scores when passing the Unified State Exam in mathematics.

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