This is the name of equations of the form where the unknown is located both in the exponent and in the base of the degree.

You can indicate a completely clear algorithm for solving an equation of the form. For this, it is necessary to pay attention to the fact that for Oh) not equal to zero, one and minus one, equality of degrees with the same bases (be it positive or negative) is possible only if the indicators are equal That is, all the roots of the equation will be the roots of the equation f (x) = g (x) The converse statement is not true, for Oh)< 0 and fractional values f (x) and g (x) expressions Oh) f (x) and

Oh) g (x) lose their meaning. That is, when going from to f (x) = g (x)(for and, extraneous roots may appear, which must be eliminated by checking against the original equation. a = 0, a = 1, a = -1 must be considered separately.

So, for a complete solution of the equation, we consider the cases:

a (x) = O f (x) and g (x) are positive numbers, then this is the solution. Otherwise, no

a (x) = 1... The roots of this equation are also the roots of the original equation.

a (x) = -1... If for a value of x satisfying this equation, f (x) and g (x) are integers of the same parity (either both are even, or both are odd), then this is the solution. Otherwise, no

For and, we solve the equation f (x) = g (x) and substituting the results obtained into the original equation, we cut off extraneous roots.

Examples of solving exponential equations.

Example # 1.

1) x - 3 = 0, x = 3.because 3> 0, and 3 2> 0, then x 1 = 3 is the solution.

2) x - 3 = 1, x 2 = 4.

3) x - 3 = -1, x = 2. Both exponents are even. This solution is x 3 = 1.

4) x - 3? 0 and x? ± 1.x = x 2, x = 0 or x = 1. For x = 0, (-3) 0 = (-3) 0 -this solution is correct x 4 = 0. For x = 1, (-2) 1 = (-2) 1 - this solution is correct x 5 = 1.

Answer: 0, 1, 2, 3, 4.

Example # 2.

By the definition of the arithmetic square root: x - 1? 0, x? one.

1) x - 1 = 0 or x = 1, = 0, 0 0 is not a solution.

2) x - 1 = 1 x 1 = 2.

3) x - 1 = -1 x 2 = 0 does not fit in the ODZ.

D = (-2) - 4 * 1 * 5 = 4 - 20 = -16 - no roots.

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

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

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

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

Some of them may seem more complicated to you, some - on the contrary, too simple. But all of them are united by one important feature: in their notation there is an exponential function $ f \ left (x \ right) = ((a) ^ (x)) $. Thus, we introduce the definition:

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

Oh well. We figured out the definition. Now the question is: how to solve all this crap? The answer is both simple and complex.

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

But there is also bad news: sometimes the authors of problems for all kinds of textbooks and exams are "inspired", and their brain inflamed with drugs begins to issue such atrocious equations that solving them becomes problematic not only for students - even many teachers get stuck on such problems.

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

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

Let's look at the following equation:

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

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

Finally, only a select few surmise that these facts can be combined and, at the output, get the following result:

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

Thus, our original equation will be rewritten as follows:

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

But this is already quite solvable! On the left in the equation there is an exponential function, on the right in the equation there is an exponential function, there is nothing else but them anywhere else. Therefore, you can "discard" the bases and stupidly equate the indicators:

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

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

If you do not understand what was happening in the last four lines, be sure to return to the topic "linear equations" and repeat it. Because without a clear understanding of this topic, it's too early for you to tackle the exponential equations.

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

Well, how to solve this? First thought: $ 9 = 3 \ cdot 3 = ((3) ^ (2)) $, so the original equation can be rewritten like this:

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

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

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

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

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

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

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

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

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

This knowledge will help us many times when we have to solve more complex problems. In the meantime, enough lyrics - it's time to study the basic algorithm for solving exponential equations.

How to solve exponential equations

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

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

According to the "naive" algorithm, according to which we acted earlier, it is necessary to represent the number $ b $ as a power of the number $ a $:

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

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

And oddly enough, this scheme works about 90% of the time. And then what about the remaining 10%? The remaining 10% are slightly "schizophrenic" exponential equations of the form:

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

Well, to what degree should 2 be raised to get 3? First? But no: $ ((2) ^ (1)) = 2 $ - not enough. Second? Also not: $ ((2) ^ (2)) = 4 $ - a bit too much. Which one then?

Knowledgeable students have probably already guessed: in such cases, when it is impossible to solve “beautifully”, “heavy artillery” - logarithms - is involved in the matter. Let me remind you that using logarithms, any positive number can be represented as a power of any other positive number (except for one):

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

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

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

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

We got a slightly strange answer: $ x = ((\ log) _ (2)) 3 $. In some other assignment, with such an answer, many would have doubted and began to double-check their solution: what if there was an error somewhere somewhere? I hasten to please you: there is no mistake here, and logarithms at the roots of exponential equations are quite a typical situation. So get used to it. :)

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

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

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

We introduced the multiplier to the logarithm argument. But no one bothers us to introduce this factor into the base:

Moreover, all three options are correct - they are just different forms of writing the same number. Which one to choose and write down in this solution is up to you.

Thus, we have learned to solve any exponential equations of the form $ ((a) ^ (x)) = b $, where the numbers $ a $ and $ b $ are strictly positive. However, the harsh reality of our world is such that such simple tasks will come across you very, very rarely. Much more often you will come across something like this:

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

Well, how to solve this? Can this be solved at all? And if so, how?

Don't panic. All these equations quickly and easily reduce to those simple formulas that we have already considered. You just need to know to remember a couple of techniques from the algebra course. And of course, there is nowhere without rules for working with degrees. I'll tell you about all this now. :)

Converting exponential equations

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

1. Write down the original equation. For example: $ ((4) ^ (x)) + ((4) ^ (x-1)) = ((4) ^ (x + 1)) - 11 $;
2. Make some kind of incomprehensible crap. Or even a few crap called "transform the equation";
3. At the output, get the simplest expressions like $ ((4) ^ (x)) = 4 $ or something else like that. Moreover, one original equation can give several such expressions at once.

With the first point, everything is clear - even my cat can write the equation on a piece of paper. With the third point, too, it seems, it is more or less clear - we have already solved a whole bunch of such equations above.

But what about the second point? What kind of transformation? What to convert to what? And How?

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

1. The equation is composed of exponential functions with the same base. Example: $ ((4) ^ (x)) + ((4) ^ (x-1)) = ((4) ^ (x + 1)) - 11 $;
2. The formula contains exponential functions with different bases. Examples: $ ((7) ^ (x + 6)) \ cdot ((3) ^ (x + 6)) = ((21) ^ (3x)) $ and $ ((100) ^ (x-1) ) \ cdot ((2.7) ^ (1-x)) = 0.09 $.

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

Highlighting a stable expression

Let's take another look at this equation:

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

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

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

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

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

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

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

The first four terms contain the element $ ((4) ^ (x)) $ - let's take it outside the parenthesis:

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

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

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

That's all! We reduced the original equation to the simplest one and got the final answer.

At the same time, in the process of solving, we found (and even took out of the bracket) the common factor $ ((4) ^ (x)) $ - this is the stable expression. It can be designated as a new variable, or it can simply be accurately expressed and answered. In any case, the key principle of the solution is as follows:

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

The good news is that virtually every exponential equation allows for such a stable expression.

But the bad news is that expressions like these can be tricky and can be tricky to pick out. Therefore, we will analyze one more task:

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

Perhaps someone will now have a question: “Pasha, are you stoned? There are different bases here - 5 and 0.2 ”. But let's try to convert the degree from base 0.2. For example, let's get rid of the decimal fraction, bringing it to the usual one:

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

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

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

Here I, of course, cheated a little. Because for a complete understanding, the formula for getting rid of negative indicators had to be written as follows:

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

On the other hand, nothing prevented us from working with only one fraction:

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

But in this case, you need to be able to raise the degree to another degree (remember: in this case, the indicators add up). But there was no need to "turn over" the fractions - perhaps for some it will be easier. :)

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

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

So it turns out that the original equation is even easier to solve than the previously considered one: here you don't even need to single out a stable expression - everything has been reduced by itself. It remains only to remember that $ 1 = ((5) ^ (0)) $, whence we get:

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

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

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

Now let's move on to more complex equations, in which there are different bases, which are generally not reducible to each other using powers.

Using the degree property

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

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

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

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

Let's start with the first equation:

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

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

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

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

Now let's deal with the second equation. Everything is much more complicated here:

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

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

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

Unfortunately, nothing really appeared in our country. But we see that the exponents on the left in the product are opposite:

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

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

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

Now note that the numbers on the left (at the bottom) and on the right are somewhat similar. How? But it is obvious: they are powers of the same number! We have:

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

Thus, our equation will be rewritten as follows:

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

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

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

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

Finally, our equation will take the form:

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

That's the whole solution. Its main idea boils down to the fact that even with different grounds, we try by hook or crook to reduce these grounds to one and the same. In this we are helped by elementary transformations of equations and rules for working with degrees.

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

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

And in order to help you in this difficult task, I suggest downloading a set of equations for your own solution on my website. All equations have answers, so you can always test yourself.

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

The solution of most mathematical problems in one way or another is associated with the transformation of numerical, algebraic or functional expressions. This applies in particular to the solution. In versions of the exam in mathematics, this type of problem includes, in particular, problem C3. Learning how to solve C3 tasks is important not only for the purpose of successfully passing the exam, but also for the reason that this skill will come in handy when studying a mathematics course in high school.

Performing tasks C3, you have to solve various types of equations and inequalities. Among them are rational, irrational, exponential, logarithmic, trigonometric, containing modules (absolute values), as well as combined ones. This article discusses the main types of exponential equations and inequalities, as well as various methods for solving them. Read about the solution of other types of equations and inequalities in the heading "" in articles devoted to methods for solving problems C3 from the USE versions in mathematics.

Before proceeding with the analysis of specific exponential equations and inequalities As a math tutor, I suggest you brush up on some of the theoretical material we need.

Exponential function

What is an exponential function?

View function y = a x, where a> 0 and a≠ 1 is called exponential function.

The main exponential function properties y = a x:

Exponential function graph

The exponential function graph is exhibitor:

Exponential function plots (exponentials)

Solving exponential equations

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

For solutions exponential equations you need to know and be able to use the following simple theorem:

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

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

Title = "(! LANG: Rendered by QuickLaTeX.com">!}

Example 1. Solve the equation:

Solution: we use the above formulas and substitution:

The equation then takes the form:

The discriminant of the resulting quadratic equation is positive:

Title = "(! LANG: Rendered by QuickLaTeX.com">!}

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

Passing to the reverse substitution, we get:

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

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

Answer: x = 3.

Example 2. Solve the equation:

Solution: the equation has no restrictions on the range of admissible values, since the radical expression makes sense for any value x(exponential function y = 9 4 -x is positive and nonzero).

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

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

Answer:x= 6.

Example 3. Solve the equation:

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

Answer: x = 0.

Example 4. Solve the equation:

Solution: we simplify the equation to an elementary one by equivalent transformations using the rules for dividing and multiplying powers given at the beginning of the article:

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

Answer: x = 0.

Example 5. Solve the equation:

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

Answer: x = -1.

Example 6. Solve the equation:

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

Answer: x = 2.

Solution of exponential inequalities

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

For solutions exponential inequalities knowledge of the following theorem is required:

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

Example 7. Solve the inequality:

Solution: we represent the original inequality in the form:

Divide both sides of this inequality by 3 2 x, moreover (due to the positivity of the function y= 3 2x) the inequality sign will not change:

Let's use the substitution:

Then the inequality takes the form:

So, the solution to the inequality is the interval:

passing to the reverse substitution, we get:

The left inequality, due to the positiveness of the exponential function, is performed automatically. Using the well-known property of the logarithm, we pass to the equivalent inequality:

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

So, we finally get answer:

Example 8. Solve the inequality:

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

Let's introduce a new variable:

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

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

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

Then, passing to the inverse substitution, we get:

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

Finally we get answer:

Example 9. Solve the inequality:

Solution:

We divide both sides of the inequality by the expression:

It is always greater than zero (due to the positiveness of the exponential function), so the sign of the inequality does not need to be changed. We get:

t located in the interval:

Passing to the inverse substitution, we find that the original inequality splits into two cases:

The first inequality of solutions does not have due to the positiveness of the exponential function. We solve the second:

Example 10. Solve the inequality:

Solution:

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

Parabola branches y = x 2 -2x+2, standing in the indicator, are directed upwards, which means it is bounded from below by the value that it reaches at its top:

At the same time, the function y = 3 x 2 -2x+2 on the right side of the equation. It reaches its smallest value at the same point as the parabola in the exponent, and this value is equal to 3 1 = 3. So, the original inequality can turn out to be true only if the function on the left and the function on the right take the value at one point equal to 3 (only this number is the intersection of the ranges of values ​​of these functions). This condition is fulfilled at a single point x = 1.

Answer: x= 1.

To learn to solve exponential equations and inequalities, it is necessary to constantly train in solving them. In this difficult matter, you can be helped by various teaching aids, problem books in elementary mathematics, collections of competition problems, mathematics classes at school, as well as individual lessons with a professional tutor. I sincerely wish you success in your preparation and excellent results on the exam.

Sergey Valerievich

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

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 shows that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of training, need to thoroughly master the theory, memorize formulas and understand the principle of solving such equations. Having learned how to cope with this type of problems, graduates will be able to count on high scores when passing the exam in mathematics.

Get ready for the exam testing with Shkolkovo!

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

The educational portal "Shkolkovo" invites students to use our knowledge base. We are implementing a completely new method of preparing for final testing. By studying on our website, you will be able to identify gaps in knowledge and pay attention to precisely those tasks that cause the greatest difficulties.

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

Basic definitions and formulas are presented in the "Theoretical Reference" section.

For better assimilation of the material, we recommend practicing the tasks. Carefully review the examples of exponential equations with a solution presented on this page to understand the calculation algorithm. After that, proceed to the tasks in the "Directories" section. You can start with the easiest problems or go straight to solving complex exponential equations with several unknowns or. The exercise base on our website is constantly being supplemented and updated.

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

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

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 the 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 the 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, you will also find a secret level - 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, 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)! Thrown 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, exactly any!) Exponential equation: using identical transformations, it is necessary to ensure that the left and right in the equation are 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: it is possible to remove identical bases if and only if in the equation on the left and right 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 rewritten in general form 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 do we so easily and simply discard the same bases on the left and right and equate 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 throw out 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 the exponential function and its properties that underlie the solution of exponential equations, yes.) A detailed answer to this question will be given in a separate special lesson 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 task at the moment is learn to solve exponential equations! The most, the simplest! Therefore - until we take a steam bath and boldly throw out the same bases. This is 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 people can at least 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 an outright 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 do not just put them.

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 relatives.) Therefore, we have every right to write:

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

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

(a m) n = a mn

If you now run it, then in general it turns out great:

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

The original example now looks like this:

3 2 x - 3 3 (x +2) = 0

Great, the bottoms of the degrees have leveled out. Which is what we wanted. Half the battle is done.) Now we start 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 not the most popular. And in the same way, by the way. That is why in exponential equations, observation and the ability to recognize powers of other numbers in numbers are so important!

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 what 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 come in handy not only at this level, but also at the next, 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, of course):

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

Yes Yes! 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 scary equation:

Again, the first glance - 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?

It's OK! 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 the usual 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 you (suddenly, who does not know) that the basic rules of actions with degrees 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. The elementary mathematics of the middle classes remains - 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. It is much easier to recognize the powers of many popular numbers in fractions! After recognition, we pass from fractions to powers with negative exponents.

Keep in mind that such a trick occurs very, very often in exponential equations! 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! It looks like a quiet horror ... However, looks are deceiving. This is the simplest exponential equation, in spite of its intimidating 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? And who knows ...

Well, again we are climbing 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 has not understood 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 everywhere to reach the same base (two), 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

Or

Here only practice and rich experience rule, yes. And, of course, a clear idea, what is negative and fractional degree. And also 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. It's not for nothing that I write in green 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 deciphering negative and fractional degrees are not in vain, I propose to play a little game!

Imagine the numbers as a power of two:

Answers (in disarray):

Happened? Excellent! 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 is 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? Excellent! 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 decided? Well, respect! Hats off.) So, the lesson was not in vain, and the initial level of solving exponential equations can be considered successfully mastered. More levels and more challenging equations are ahead! And new techniques and approaches. And non-standard examples. And new surprises.) All this is in the next lesson!

Did something go wrong? This means, 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.)