Exponential equations in two unknowns. Exponential equations. How to solve exponential equations

What is an exponential equation? Examples.

So, an exponential equation ... A new unique exhibit at our common exhibition of a wide variety of equations!) As it almost always happens, 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 in terms of any degree. And only there! This is extremely important.

For example, such simple equations:

3 x +1 = 81

5 x + 5 x +2 = 130

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

Or even monsters like this:

2 sin x = 0.5

I ask you to immediately pay attention to one important thing: in grounds degrees (bottom) - only numbers... But in indicators degrees (top) - a wide variety of expressions with x. Absolutely any.) Everything depends on the specific equation. If, suddenly, x appears in the equation somewhere else, 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. Therefore, we will not consider them in this lesson. To the delight of the students.) Here we will consider only exponential equations in a "pure" form.

Generally speaking, even pure exponential equations are far from being solved clearly and not always. 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 will definitely solve examples.) So let's get comfortable and - off we go! As in computer shooters, our journey will take place through the levels.) From elementary to simple, from simple to intermediate and from intermediate to difficult. On the way, a secret level will also be waiting for you - techniques and methods for solving non-standard examples. Those that you won't read about in most school textbooks ... Well, at the end, of course, there is a final boss in the form of homework.)

Level 0. What is the simplest exponential equation? Solution of the simplest exponential equations.

To begin with, let's consider some frank elementary stuff. You have to start somewhere, right? For example, an equation like this:

2 x = 2 2

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

2 x = 2 2

X = 2

What happened with us? And the following happened. We, in fact, took and ... just threw out the same bases (deuces)! They threw it out completely. And, what pleases, hit the bull's-eye!

Yes, indeed, if the exponential equation on the left and right contains the same numbers in any powers, then these numbers can be discarded and simply equate the exponents. Mathematics solves.) And then you can work separately with the indicators and solve a much simpler equation. Great, isn't it?

This is the key idea of solving any (yes, any!) Exponential equation: via identical transformations it is necessary to ensure that the left and right in the equation stand the same base numbers in varying degrees. And then you can safely remove the same bases and equate the degree indicators. And work with a simpler equation.

And now we remember the iron rule: removing the same bases is possible if and only if in the equation on the left and right of the base numbers are in proud loneliness.

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

For example, in the equation

3 3 x-5 = 3 2 x +1

You cannot remove the triplets! Why? Because on the left we have not just a lonely three in degree, but work 3 3 x-5. The extra three gets in the way: the coefficient, you know.)

The same can be said about the equation

5 3 x = 5 2 x +5 x

Here, too, all bases are the same - five. But on the right we have not a lone degree of five: there is the sum of the degrees!

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

af (x) = a g (x)

This type of exponential equation is called the simplest... Or, scientifically, canonical ... And whatever twisted equation we have in front of us, we, one way or another, will reduce it to this very simple (canonical) form. Or, in some cases, to the aggregate equations of this kind. Then our simplest equation can be in general view rewrite like this:

F (x) = g (x)

And that's all. This will be the equivalent conversion. In this case, absolutely any expressions with an x can be used as f (x) and g (x). Anything.

Perhaps a particularly inquisitive student will ask: why on earth are we so easily and simply discarding the same bases on the left and right and equating the degree indicators? Intuition by intuition, but suddenly, in some equation and for some reason, this approach turns out to be wrong? Is it always legal to discard the same grounds? Unfortunately, for a rigorous mathematical answer to this interesting question, one needs to plunge rather deeply and seriously into the general theory of the structure and behavior of functions. And a little more specifically - into a phenomenon strict monotony. In particular, the strict monotonicity exponential functiony= a x... Since it is exponential function and its properties underlie the solution of exponential equations, yes.) A detailed answer to this question will be given in a separate special lesson devoted to solving complex non-standard equations using the monotonicity of different functions.)

Explaining this moment in detail now is only to take out the brain of the average schoolboy and scare him off prematurely with a dry and heavy theory. I will not do this.) For our main one is this moment task - learn to solve exponential equations! The most, the simplest! Therefore - until we take a steam bath and boldly throw out the same bases. This can, take my word for it!) And then we solve the equivalent equation f (x) = g (x). Typically simpler than the original indicative.

It is assumed, of course, that at least people can solve the equations, already without x in the indicators, at the moment.) Who still does not know how - feel free to close this page, follow the corresponding links and fill in the old gaps. Otherwise, you will have a hard time, yes ...

I am already silent about the irrational, trigonometric and other brutal equations, which can also emerge in the process of eliminating the grounds. But do not be alarmed, we are not going to consider frank tin in terms of degrees: it is too early. We will train only on the simplest equations.)

Now let's look at equations that require some extra effort to reduce them to the simplest ones. 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. We recognize the degrees! Natural indicators.

The key rules in solving any exponential equations are power rules... Without this knowledge and skills, nothing will work. Alas. So, if with the degrees of the problem, then first you are welcome. In addition, we will need more. These transformations (as many as two!) Are the basis for solving all equations of mathematics in general. And not only indicative. So, who have forgotten, also take a walk on the link: I put them for a reason.

But actions with degrees and identical transformations alone are not enough. You also need personal observation and ingenuity. 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, an equation like this:

3 2 x - 27 x +2 = 0

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

27 = 3 3

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

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

And now we connect our knowledge about actions with degrees(and I warned!). There is a very useful formula there:

(a m) n = a mn

If you now start it up, then it works out great in general:

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 bottoms of the degrees have leveled out. Which is what we wanted. Half the battle has been done.) And now we launch the basic identity transformation - move 3 3 (x +2) to the right. Nobody canceled the elementary actions of mathematics, yes.) We get:

3 2 x = 3 3 (x +2)

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

2x = 3 (x + 2)

We solve this and get:

X = -6

That's all there is to it. This is the correct answer.)

And now we comprehend the course of the decision. What saved us in this example? We were saved by the knowledge of the degrees of the three. How exactly? We identified among 27 encrypted three! This trick (encrypting the same base under different numbers) is one of the most popular in exponential equations! If only not the most popular. And in the same way, by the way. That is why observation and the ability to recognize powers of other numbers in exponential equations are so important in exponential equations!

Practical advice:

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

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

Since the ability to recognize degrees in the face will be useful not only at this level, but also at the following, here's a little task for you:

Determine what powers and what numbers are numbers:

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

Answers (randomly, naturally):

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! Do not 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. We 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 build up the power, right?

For example, this terrible equation:

Again the first glance is at the foundations. The grounds are different! And this time, even remotely different from each other! 5 and 0.04 ... And to eliminate the grounds, you need the same ... What to do?

Nothing wrong! In fact, everything is the same, just the connection between the five and 0.04 is visually poorly visible. How do we get out? And let's move on in the number 0.04 to ordinary fraction! And there, you see, everything will be formed.)

0,04 = 4/100 = 1/25

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

How is it? Is it easier to see the relationship between 5 and 1/25 now? That's it ...

And now, according to the rules of action with powers with negative indicator you can write down with a firm hand:

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

Again, according to the rules for dealing with powers, you can now write:

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

Just in case, I remind (suddenly, who does not know) that basic rules actions with powers are valid for any indicators! Including for negative ones.) So we can safely take and multiply the indicators (-2) and (x-1) according to the appropriate rule. Our equation keeps getting better and better:

Everything! Apart from the lonely fives in the degrees on the left and on the right, there is nothing else. The equation is reduced to the 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. There remains elementary mathematics of the middle classes - we open (right!) 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 to varying degrees! Namely, to see the encrypted five in the number 0.04. And this time - in negative degree! How did we do it? On the move - nothing. But after the transition from a decimal fraction of 0.04 to an ordinary fraction of 1/25, everything was highlighted! And then the whole decision went like clockwork.)

Therefore, another green practical advice.

If decimal fractions are present in the exponential equation, then we go from decimal fractions to ordinary ones. V ordinary fractions it is much easier to recognize the powers of many popular numbers! After recognition, we pass from fractions to powers with negative exponents.

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

Solve the equation:

In! In appearance - a quiet horror ... However, appearances are deceiving. This is the simplest exponential equation, despite its daunting appearance... And now I’ll show you.)

First, we deal with all the numbers sitting in the bases and in the coefficients. They are, of course, different, yes. But we still take the risk and try to make them the same! Let's try to get to the same number in different degrees... And, preferably, the number of the smallest possible. So, let's start decrypting!

Well, with a four, everything is clear at once - it's 2 2. So, already something.)

With a fraction of 0.25 - it is not yet clear. It is necessary to check. We use a practical advice - we go from decimal fraction to ordinary one:

0,25 = 25/100 = 1/4

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

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

Well, once again we climb into our treasury of knowledge about degrees! This time we additionally connect our knowledge about the 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 exponent.

Like this:

In our case:

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

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

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

Who did not understand exactly how this equation came about, then the question 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 home stretch! The canonical form of the exponential equation is obtained! How is it? Have I convinced you that everything is not so scary? ;) We remove the deuces and equate the indicators:

All that remains is to solve this linear equation. How? With the help of identical transformations, obviously.) Make it up, what is already there! Multiply both parts by two (to remove the fraction 3/2), transfer terms with x to the left, without x to the right, bring similar ones, count - and you will be happy!

Everything should turn out beautifully:

X = 4

And now we again comprehend the course of the decision. In this example, we were helped out by the transition from square root To degree with exponent 1/2... Moreover, only such a cunning transformation helped us to 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 another practical advice:

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

Of course, negative and fractional degrees are already much more complicated than 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 -3 power or four to the -3/2 power is not such a big problem. For those in the know.)

But go, for example, figure out right away that

0,125 = 2 -3

Here only practice and rich experience rule, yes. And, of course, a clear idea, what is negative and fractional degree. As well as - practical advice! Yes, yes, those green.) I hope that they will still help you to better navigate in all the motley variety of degrees and will significantly increase your chances of success! So do not neglect them. I'm not in vain green I write sometimes.)

But if you become familiar even with such exotic degrees as negative and fractional, then your possibilities in solving exponential equations will expand enormously, and you will already be able to handle almost any type of exponential equations. Well, if not any, then 80 percent of all exponential equations - for sure! Yes, I’m not kidding!

So, our first part of getting to know the exponential equations has come to its logical conclusion. And, as an intermediate workout, I traditionally suggest doing a little on your own.)

Exercise 1.

So that my words about decoding negative and fractional powers did not go to waste, I propose to play a little game!

Imagine the numbers as a power of two:

Answers (in disarray):

Happened? Fine! Then we do a combat mission - we solve the simplest and simplest exponential equations!

Task 2.

Solve equations (all answers are in disarray!):

5 2x-8 = 25

2 5x-4 - 16 x + 3 = 0

Answers:

x = 16

x 1 = -1; x 2 = 2

x = 5

Happened? Indeed, it’s much easier!

Then we solve the following 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? Fine! You are growing! Then here are some more examples for a snack:

Answers:

x = 6

x = 13/31

x = -0,75

x 1 = 1; x 2 = 8/3

And is it settled? Well, respect! I take off my hat.) So the lesson was not in vain, and First level solutions of exponential equations can be considered successfully mastered. Ahead - more levels and more difficult equations! 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, most likely, problems in. Or in . Or both at once. Here I am powerless. I can once again offer only one thing - not to be lazy and take a walk through the links.)

To be continued.)

Exponential equations. As you know, the USE includes simple equations. We have already considered some of them - these are logarithmic, trigonometric, rational. Here are indicative equations.

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

Let's list the properties of exponents:

The zero degree of any number is equal to one.

Consequence of this property:

A little more theory.

An exponential equation is an equation containing a variable in an indicator, that is, this equation of the form:

f(x) an 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. When obtaining an equation of the form a f (x) = b the definition of the logarithm is used, we get:

3. As a result of transformations, an equation of the form can be obtained:

The logarithm is applied:

We express and find x.

In tasks options for the exam it will be enough to use the first method.

That is, it is necessary to present the left and right parts in the form of degrees with the same base, and then we equate the indicators and solve the usual linear equation.

Consider the equations:

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

It is necessary to make sure that on the left and right sides there are indicative expressions with one base. 64 we can represent 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:

We 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 represent 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 degree we raise a positive number a, we cannot get a negative number in any way.

Any exponential equation after appropriate transformations is reduced to the solution of one or more of the simplest ones.In this section, we will also look at the solution of some equations, do not miss it!That's all. Success to you!

Best regards, Alexander Krutitskikh.

P.S: I would be grateful if you could tell us 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 reduce to the form \ (a ^ (f (x)) = a ^ (g (x)) \), and then make the transition to equality of indicators, that is:

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

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

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

For instance:

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 \). With this in mind, we 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 \ (\ sqrt (3 ^ 3) = ((3 ^ 3)) ^ ( \ frac (1) (2)) \). Further, using the degree property \ ((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 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-hand side, we get: \ ((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. This means 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 property of degree \ (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 the degree, 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 closely at the equation, and we see that the replacement \ (t = 2 ^ x \) suggests itself.

\ (t_1 = 2 \) \ (t_2 = \ frac (1) (2) \)		However, we found the values \ (t \), but we need \ (x \). We return to the Xs, making the reverse replacement.
\ (2 ^ x = 2 \) \ (2 ^ x = \ frac (1) (2) \)		Transform the second equation using the negative exponent property ...
\ (2 ^ x = 2 ^ 1 \) \ (2 ^ x = 2 ^ (- 1) \)		... and we decide to answer.
\ (x_1 = 1 \) \ (x_2 = -1 \)

Answer : \(-1; 1\).

The question remains - how to understand when to apply which method? It comes with experience. Until you get it, use general recommendation to solve complex problems - “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 - suddenly what happens? The main thing is to do only mathematically justified transformations.

Exponential Equations Without Solutions

Let's look at two more situations that often baffle students:
- a positive number to a power is equal to zero, for example, \ (2 ^ x = 0 \);
- a positive number to the power is negative number, for example, \ (2 ^ x = -4 \).

Let's try to solve it by brute force. If x is a positive number, then as x grows, the entire power \ (2 ^ x \) will only grow:

\ (x = 1 \); \ (2 ^ 1 = 2 \)
\ (x = 2 \); \ (2 ^ 2 = 4 \)
\ (x = 3 \); \ (2 ^ 3 = 8 \).

\ (x = 0 \); \ (2 ^ 0 = 1 \)

Also by. There are negative x's left. Remembering 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 gets smaller with each step, it will never reach zero. So the negative degree did not save us either. We come to a logical conclusion:

A positive number will remain positive to any extent.

Thus, both equations above have no solutions.

Exponential equations with different bases

In practice, sometimes there are 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 instance:

\ (7 ^ (x) = 11 ^ (x) \)
\ (5 ^ (x + 2) = 3 ^ (x + 2) \)
\ (15 ^ (2x-1) = (\ frac (1) (7)) ^ (2x-1) \)

Such equations can be easily solved by dividing by any of the parts of the equation (usually divided by the right side, that is, \ (b ^ (f (x)) \). You can divide this way, because a positive number is positive to any degree (that is, we do not divide by zero). We get:

\ (\ 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 cannot turn the five into a three, or vice versa (at least, without using it). So we cannot come to the form \ (a ^ (f (x)) = a ^ (g (x)) \). In this case, the indicators are the same. Let's divide the equation by the right-hand side, that is, by \ (3 ^ (x + 7) \) (we can do this, since we know that the triple is not zero in any way).
\ (\ frac (5 ^ (x + 7)) (3 ^ (x + 7)) \) \ (= \) \ (\ frac (3 ^ (x + 7)) (3 ^ (x + 7) ) \)		Now we recall 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 it did not get better. But remember one more property of the degree: \ (a ^ 0 = 1 \), in other words: "any number in the zero degree is equal to \ (1 \)". The converse is also true: "one can be represented as any number to the zero degree." We use this by making the base on the right the same as on the left.
\ ((\ frac (5) (3)) ^ (x + 7) \) \ (= \) \ ((\ frac (5) (3)) ^ 0 \)		Voila! We get rid of the bases.
		We write the answer.

Answer : \(-7\).

Sometimes the "similarity" of the exponents is not obvious, but the skillful use of the properties of the degree solves 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 quite sad ... Not only can the bases not be reduced to the same number (the seven will not be equal to \ (\ frac (1) (3) \)), but also the indicators are different ... However, let's in the left exponent two.
\ (7 ^ (2 (x-2)) = (\ frac (1) (3)) ^ (- x + 2) \)		Remembering the property \ ((a ^ b) ^ c = a ^ (b c) \), 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, recalling the property of negative degree \ (a ^ (- n) = \ frac (1) (a) ^ n \), 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 have become the same! Acting according to the scheme already familiar to us, we decide before answering.

Answer : \(2\).

First level

Exponential equations. Comprehensive guide (2019)

Hey! Today we will discuss with you how to solve equations, which can be both elementary (and I hope that after reading this article, almost all of them will be for you), and those that are usually given "for filling." Apparently, to fall asleep completely. But I will try to do my best so that now you do not get screwed up when faced with this type of equations. I will not beat around the bush anymore, but I will immediately open little secret: today we will be engaged exponential equations.

Before proceeding to the analysis of ways to solve them, I will immediately outline in front of you a circle of questions (rather small), which you should repeat before rushing to storm this topic. So, for the best result, please repeat:

Properties and
Solution and equations

Repeated? Wonderful! 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? Truth? Then let's continue. Now answer me the question, what is the third degree? You're absolutely right: . And eight is what a power of two? That's right - the third! Because. Well, now let's try to solve the following problem: Let me multiply the number by myself once and get the result. The question is, how many times have I multiplied 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 by myself I have multiplied times. How else can you check? And here's how: directly by definition of the degree:. But, you must admit, if I asked how many times two must be multiplied by itself in order to get, say, you would have told me: I will not fool myself and multiply by myself to the point of blue in the face. And he would be absolutely right. Because how can you write down all actions briefly(and brevity is the sister of talent)

where - these are the same "Times" when you multiply by yourself.

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:

Where can you make a fully justified conclusion that:

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

And even found him root... Don't you think that everything is completely trivial? So I think exactly the same. Here's another example for you:

But what is to be done? You can't write it down as a power of a (reasonable) number. Let's not despair and note that both of these numbers are perfectly expressed in terms of 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 pull any more and write definition:

In our case with you:.

These equations are solved by reducing them to the form:

with the subsequent solution of the equation

We, in fact, did this in the previous example: we got that. And we solved the simplest equation with you.

It seems to be nothing complicated, right? Let's practice the simplest ones first. examples:

We again see that the right and left sides of the equation need to be represented as a power of one number. True, this has already been done on the left, but there is a number on the right. But, it's okay, because my equation will miraculously transform into this:

What did I have to use here? What is the rule? Degree to Degree Rule which reads:

What if:

Before answering this question, let's fill in the following plate:

It is not difficult for us to notice that the less, the less value but nonetheless, all of these values are greater than zero. AND THIS WILL ALWAYS BE !!! The same property is true FOR ANY BASE WITH ANY INDICATOR !! (for any and). Then what can we conclude about the equation? And here's what: it has no roots! Neither does any equation have roots. Now let's practice and Let's solve simple examples:

Let's check:

1. Nothing is required from you here, except for the knowledge of the properties of the degrees (which, by the way, I asked you to repeat!) As a rule, everything leads to the least reason:,. Then the original equation is equivalent to the following: All I need is to use the properties of the degrees: when multiplying numbers with the same bases, the powers are added, and when dividing, they are subtracted. Then I get: Well, now, with a clear conscience, I will move from an exponential equation to a 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, you need to be more careful: the trouble is that on the left side we will not be able to present it in the form of a power of the same number. In this case, it is sometimes useful represent numbers as a product of degrees with different bases, but the same indicators:

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 indicators can be multiplied.In this case, the bases are multiplied, and the indicator does not change:

Applied to 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 do not like it when, unnecessarily, on the one side of the equation there are two terms, and on the other - none (sometimes, of course, this is justified, but this is not the case now). Move the minus term to the right:

Now, as before, I will write everything in terms of powers of a triple:

Add the powers to the left and get the equivalent equation

You can easily find its root:

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

On the left, I'm almost all right, except for what? Yes, the "wrong degree" in the deuce bothers me. But I can easily fix it by writing:. Eureka - on the left, all bases are different, but all degrees are the same! Multiply urgently!

Here, again, everything is clear: (if you did not understand how magically I got the last equality, take a break for a minute, take a break 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 that 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 tasks for you to train, to which I will only give the answers (but in a "mixed" form). Cut them down, check them, and you and I will continue our research!

Ready? Answers like these ones:

any number

Okay, okay, I was joking! Here's an outline of the solutions (some are very short!)

Don't you think it is no coincidence that one fraction on the left is an “inverted” other? 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 be like this:

Having decided this quadratic equation, you will get these roots:

2. Another solution: dividing both sides of the equation by the expression on the left (or right). I divide by what is on the right, then I get:

Where (why ?!)

3. I do not even want to repeat myself, everything is already "chewed up" so much.

4.equal to a quadratic equation, roots

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

The equation has become a trivial identity, which is true for any. Then the answer is any real number.

Well, so you've practiced solving the simplest exponential equations. Now I want to give you a few life examples that will help you understand why they are needed in principle. I will give two examples here. One of them is quite everyday, but the other is more likely to be of scientific rather than practical interest.

Example 1 (mercantile) Suppose you have rubles, and you want to turn it into rubles. The bank offers you to take this money from you at an annual interest rate with monthly capitalization of interest (monthly accrual). The question is, for how many months do you need to open a deposit in order to collect 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, - the interest rate for the period, - the number of periods. Then:

In our case (if the rate is per annum, then it is charged per month). Why is it divided by? If you do not know the answer to this question, remember the topic ""! Then we get the following equation:

This exponential equation can already be solved only with the help of a calculator (its appearance hints at this, and this requires knowledge of logarithms, which we will get to know a little later), which I will do: ... Thus, to get a million, we need to make a contribution for a month (not very fast, right?).

Example 2 (more scientific). Despite his, some "isolation", I recommend that you pay attention to him: he regularly "slips in the exam !! (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. In how many minutes will the mass of the isotope be equal to mg? It's okay: we just take and substitute all the data in the formula proposed to us:

Let's split both parts into, "in the hope" that on the left we get something digestible:

Well, we are very lucky! It stands on the left, then we turn to the equivalent equation:

Where is the min.

As you can see, the exponential equations have a very real application in practice. Now I want to discuss with you another (simple) way to solve exponential equations, which is based on taking the common factor out of the parentheses, followed by grouping the terms. Do not be intimidated by my words, you already came across this method in the 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 the squares:

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

Then the original expression is equivalent to this:

Where to take out the common factor is no longer difficult:

Hence,

This is approximately how we will act when solving exponential equations: look for "commonality" among the terms and put it outside the brackets, well then - come what may, I believe that we will be lucky =)) For example:

On the right is far from a power of seven (I checked it!) And on the left - a little better, you can, of course, "chop off" the factor a from the second from the first term, and then deal with the result, but let's do it more sensibly with you. I don’t want to deal with fractions that inevitably come from “highlighting,” so wouldn't it be better for me to endure? Then I won't have fractions: as they say, the wolves are fed and the sheep are safe:

Count the expression in parentheses. In a magical, magical way, it turns out that (surprising, although what else can we expect?).

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

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

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

Now let's move the "common" to 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 take a deeper look:

Well, now let's make it so that on the left we have only the expression with, and on the right - everything else. How do we do this? And here's how: Divide both sides of the equation first by (this way we get rid of the degree on the right), and then divide both sides by (this way we get rid of the numerical factor on the left). We finally get:

Incredible! On the left we have an expression, and on the right we have a simple one. Then we immediately conclude that

Here's another example for you to consolidate:

I will bring him short solution(not really bothering yourself with explanations), try to figure out all the "subtleties" of the solution yourself.

Now the final consolidation of the passed material. Try to solve the following problems yourself. I will only give brief recommendations and tips for solving them:

Let's take the common factor out of the brackets:
We represent the first expression in the form:, divide both parts into and get that
, then the original equation is transformed to the form: Well, now a hint - look where you and I have already solved this equation!
Imagine how, how, and, well, then divide both parts by, so you get the simplest exponential equation.
Take out of the brackets.
Take out of the brackets.

EXPLORATIVE EQUATIONS. AVERAGE LEVEL

I guess after reading the first article that told what are exponential equations and how to solve them you mastered the necessary minimum knowledge required to solve the simplest examples.

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

"Method of introducing a new variable" (or replacement). He solves most of the "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 miraculously transforms into one that you can already easily solve. All that is left for you after solving this very “simplified equation” is to make a “reverse replacement”: that is, to return from the replaced to the replaced one. Let's illustrate what we just said with a very simple example:

Example 1:

This equation is solved using "simple substitution," as mathematicians disparagingly call it. Indeed, the replacement here is the most obvious one. One has only to see that

Then the original equation turns into this:

If we additionally present how, then it is quite clear what needs to be replaced: of course,. What then will the original equation turn into? And here's what:

You can easily find its roots on your own:. What should we do now? It's time to go back to the original variable. What did I forget to indicate? Namely: when replacing a certain degree with a new variable (that is, when changing a view), 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.

Answer:

As you can see, in the previous example, the replacement was asking for our hands. Unfortunately, this is not always the case. However, let's not go straight to the sad, but practice with one more example with a fairly simple replacement

Example 2.

It is clear that most likely it will be necessary to replace (this is the smallest of the powers included in our equation), however, before introducing the replacement, our equation must be "prepared" for it, namely:,. Then you can replace, as a result I get the following expression:

Oh horror: a cubic equation with completely creepy formulas for its solution (well, speaking in general terms). But let's not despair right away, but think about what to do. I will propose to cheat: we know that in order to get a “nice” answer, we need to get it in the form of some power of a triplet (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 assumption. It is not a root. Alas and ah ...

.
The left side is equal.
Right part: !
There is! You have guessed the first root. Now things will get easier!

Do you know about the “corner” division scheme? Of course you know 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 great theorem:

Applied to my situation, this tells me what is divisible by. How is division carried out? That's how:

I look at which monomial I have to multiply to get It is clear that on, then:

Subtract the resulting expression from, 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 last step, multiply by, and subtract from the remaining expression:

Hurray, the division is over! What have we saved up in private? By itself: .

Then we got the following decomposition 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 want to scare you with this example, but rather my goal was to show that although we had a fairly simple replacement, nevertheless it led to a rather complicated 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 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 reasons and one foundation is not obtained from another by raising to any (reasonable, naturally) degree. 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, a smart move would be multiply both sides of the equation by the conjugate number.

For example, on, then the left side of the equation becomes equal, and the right side. 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 of the "school" exponential equations. The following tasks are taken from the exam C1 (advanced level of difficulty). You are already competent enough to independently solve these examples. I will only give the required replacement.

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

And now, brief explanations and answers:

Here it is enough for us to note that and. Then the original equation will be equivalent to this one: This equation is solved by replacing Further calculations do it yourself. In the end, your task will be reduced to solving the simplest trigonometric (depending on sine or cosine). We will analyze the solution of such examples in other sections.
Here you can even do without replacement: just move the subtracted to the right and represent both bases through powers of two:, and then go directly to the quadratic equation.
The third equation is also solved in a fairly standard way: let's imagine how. Then, replacing we get a quadratic equation: then,
Do you already know what a logarithm is? Not? Then read the topic urgently!

The first root, obviously, does not belong to the segment, and the second is incomprehensible! But we will find out very soon! Since, then (this is a property of the logarithm!) Compare:

Subtract from both parts, then we get:

The left side can be represented as:

multiply both parts by:

can be multiplied by, then

Then let's compare:

since, then:

Then the second root belongs to the required interval

Answer:

As you see, the selection of roots of exponential equations requires a sufficiently deep knowledge of the properties of logarithms so I advise you to be as careful as possible when solving the exponential equations. As you can imagine, in mathematics, everything is interconnected! As my math teacher used to say: "math, like history, you can't 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 quite simple to solve. By making the substitution, we will reduce our original equation to the following:

First, let's look at the first root. Compare and: since, then. (property of the logarithmic function, at). Then it is clear that the first root does not belong to our interval either. 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 and. This peg is a number. The first expression is smaller and the second is larger. Then the second expression is greater than the first and the root belongs to the interval.

Answer: .

To wrap up, let's look at another example of an equation where the replacement is rather non-standard:

Let's start right away with what you can do, and what - in principle, you can, but it's better not to do it. You can - represent everything through powers of three, two and six. Where it leads? Yes, it will not lead to anything: a hodgepodge of degrees, and some of them will be quite difficult to get rid of. What then is needed? Let's notice that And what will it give us? And the fact that we can reduce the solution this example to the solution of a simple exponential equation! First, let's rewrite our equation as:

Now we divide both sides of the resulting equation by:

Eureka! Now we can replace, we get:

Well, now it's your turn to solve demonstration problems, and I will give only brief comments to them so that you do not go astray! Good luck!

1. The most difficult! It’s not easy to find a replacement here! But nevertheless, this example can be completely solved using selection of a full square... To solve it, it is enough to note that:

Then here's a replacement for you:

(Please note that here, during our replacement, we cannot drop the negative root !!! And why do you think?)

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

Both of them are solved by the "standard replacement" (but the second in one example!)

2. Note 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 replace or.

5. Note that the numbers and are conjugate.

EXPLORATIVE EQUATIONS. ADVANCED LEVEL

In addition, let's consider another way - solution of exponential equations by the logarithm method... I cannot say that the solution of exponential equations by this method is very popular, but in some cases only it can lead us to correct decision our equation. It is especially often used to solve the so-called " mixed equations": That is, those where functions of different types meet.

For example, an equation of the form:

v general case can only be solved by taking the logarithm of both sides (for example, by the base), in which the original equation turns into the following:

Let's take a look at the following example:

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

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

As you can see, taking the logarithm of our original equation quickly enough led us to the correct (and beautiful!) Answer. Let's practice with one more example:

Here, too, there is nothing to worry about: we logarithm both sides of the equation by the base, then we get:

Let's make a replacement:

However, we are missing something! Have you noticed where I went wrong? After all, then:

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

Answer:

Try to write down the solution of the exponential equations given below yourself:

Now check your decision against this:

1. Logarithm both sides to the base, taking into account that:

(the second root does not suit us due to the replacement)

2. We logarithm to the base:

Let's transform the resulting expression to the following form:

EXPLORATIVE EQUATIONS. BRIEF DESCRIPTION AND BASIC FORMULAS

Exponential equation

Equation of the form:

called the simplest exponential equation.

Power properties

Approaches to the solution

Bringing to on the same basis
Conversion to the same exponent
Variable substitution
Simplification of expression and application of one of the above.

In this article, you will learn about all types exponential equations and algorithms for their solution, learn to recognize which type it belongs to exponential equation that you need to solve and apply the appropriate method to solve it. Detailed solution of examples exponential equations you can watch each type in the corresponding VIDEO TUTORIALS.

An exponential equation is an equation in which the unknown is contained in the exponent.

Before you start solving the exponential equation, it is helpful to do a few preliminary actions , which can greatly facilitate the course of its solution. These are the steps:

1. Factor all the bases of the powers.

2. Present the roots in the form of a degree.

3. Decimal fractions imagine as ordinary.

4. Mixed numbers write it down as irregular fractions.

You will realize the benefits of these actions in the process of solving equations.

Let's consider the main types exponential equations and algorithms for their solution.

1. Equation of the form

This equation is equivalent to the equation

Look in this VIDEO TUTORIAL to solve the equation of this type.

2. Equation of the form

In equations of this type:

b) the coefficients for the unknown in the exponent are equal.

To solve this equation, you need to bracket the factor to the smallest degree.

An example of solving this type of equation:

look in the VIDEO LESSON.

3. Equation of the form

Equations of this type differ in that

a) all degrees have the same bases

b) the coefficients for the unknown in the exponent are different.

Equations of this type are solved by changing variables. Before introducing a replacement, it is advisable to get rid of free members in the exponent. (, , etc)

Watch the VIDEO TUTORIAL for solving this type of equation:

4. Homogeneous equations of the kind

Distinctive features of homogeneous equations:

a) all monomials have the same degree,

b) the free term is zero,

c) the equation contains degrees with two different bases.

Homogeneous equations are solved using a similar algorithm.

To solve an equation of this type, divide both sides of the equation by (can be divided by or by)

Attention! When dividing the right and left sides of the equation by an expression containing the unknown, you can lose roots. Therefore, it is necessary to check if the roots of the expression by which we divide both sides of the equation are not the roots of the original equation.

In our case, since the expression is not equal to zero for any values of the unknown, we can divide by it without fear. Divide the left side of the equation by this expression term by term. We get: