So called the equations of the species where the unknown is both in the indicator and at the bottom of the degree.

You can specify a completely clear algorithm for solving the equation of the form. To do this, you need to pay attention to the fact that oh) Not equal to zero, unit and minus unit equality of degrees with the same bases (whether positive or negative) is possible only if the indicators are equal to that - there are all the roots of the equation will be roots of the equation f (x) \u003d g (x) The opposite approval is incorrectly oh)< 0 and fractional values f (x) and g (x)expressions oh) f (x) and

oh) g (x) lose meaning. Then - there is when moving from to f (x) \u003d g (x) (when foreign roots may appear, which must be excluded by testing on the source equation. And cases a \u003d 0, a \u003d 1, and \u003d -1it is necessary to consider separately.

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

a (x) \u003d o f (x) and g (x) Will with positive numbers, then this decision. Otherwise no

a (x) \u003d 1. The roots of this equation are roots and an initial equation.

a (x) \u003d -1. If, with the value of x, satisfying this equation, f (x) and g (x) These are whole numbers of the same parity (or both even, or both odd), then this decision. Otherwise no

When and solve the equation f (x) \u003d g (x) and substituting the results obtained into the initial equation, comparting foreign roots.

Examples of solving indicative-power equations.

Example number 1.

1) x - 3 \u003d 0, x \u003d 3. Because 3\u003e 0, and 3 2\u003e 0, then x 1 \u003d 3 is a solution.

2) x - 3 \u003d 1, x 2 \u003d 4.

3) x - 3 \u003d -1, x \u003d 2. Both indicators are even. This solution is x 3 \u003d 1.

4) x - 3? 0 and x? ± 1. x \u003d x 2, x \u003d 0 or x \u003d 1. with x \u003d 0, (-3) 0 \u003d (-3) 0 is a solution x 4 \u003d 0. With x \u003d 1, (-2) 1 \u003d (-2) 1 - true this solution x 5 \u003d 1.

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

Example number 2.

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

1) x - 1 \u003d 0 or x \u003d 1, \u003d 0, 0 0 This is not a solution.

2) X - 1 \u003d 1 x 1 \u003d 2.

3) x - 1 \u003d -1 x 2 \u003d 0 is not suitable in ...

D \u003d (-2) - 4 * 1 * 5 \u003d 4 - 20 \u003d -16 - no roots.

This lesson is designed for those who are just starting to study the indicative equations. As always, let's start with the definition and simplest examples.

If you read this lesson, I suspect that you already have at least a minimum idea of \u200b\u200bthe simplest equations - linear and square: $ 56x-11 \u003d 0 $; $ ((x) ^ (2)) + 5x + 4 \u003d 0 $; $ ((x) ^ (2)) - 12x + 32 \u003d 0 $, etc. To be able to solve such structures are absolutely necessary in order not to "hang" in the topic that we are talking about.

So, the indicative equations. Immediately I will give a couple of examples:

\\ [((2) ^ (x)) \u003d 4; \\ quad (((5) ^ (2x-3)) \u003d \\ FRAC (1) (25); \\ quad ((9) ^ (x)) \u003d - 3 \\]

Some of them may seem more complex, some - on the contrary, too simple. But all of them combines one important feature: in their records there is an indicative function $ F \\ left (x \\ right) \u003d ((a) ^ (x)) $. Thus, we introduce the definition:

The indicative equation is any equation containing an indicative function, i.e. Expression of the type $ ((a) ^ (x)) $. In addition to this function, such equations may contain any other algebraic designs - polynomials, roots, trigonometry, logarithms, etc.

Oh well. Defined figured out. Now the question is: how to solve all this crap? The answer is simultaneously simple, and complicated.

Let's start with good news: in your own experience, classes with many students I can say that most of them are indicative equations are much easier than the same logarithms and the more so the trigonometry.

But there is also bad news: sometimes there are "inspiration" tasks for all kinds of textbooks and exams, and their inflamed brain begins to issue such brutal equations that it becomes problematic not only to students - even many teachers stick to such tasks.

However, we will not be about sad. And back to those three equations that were presented at the very beginning of the narrative. Let's try to solve each of them.

The first equation: $ ((2) ^ (x)) \u003d $ 4. Well, what extent you need to build a number 2 to get the number 4? Probably in the second? After all, $ ((2) ^ (2)) \u003d 2 \\ Cdot 2 \u003d 4 $ - and we obtained the right numerical equality, i.e. really $ x \u003d $ 2. Well, thanks, Cap, but this equation was so simple that I would even solve my cat. :)

Let's look at the following equation:

\\ [((5) ^ (2x-3)) \u003d \\ FRAC (1) (25) \\]

And here is already a little more difficult. Many students know that $ ((5) ^ (2)) \u003d $ 25 is a multiplication table. Some also suspect that $ ((5) ^ (- 1)) \u003d \\ FRAC (1) (5) $ is essentially the definition of negative degrees (by analogy with the $ formula ((a) ^ (- n)) \u003d \\ FRAC (1) (((a) ^ (n))) $).

Finally, only the favorites guess that these facts can be combined and at the output to get the following result:

\\ [\\ FRAC (1) (25) \u003d \\ FRAC (1) (((5) ^ (2))) \u003d ((5) ^ (- 2)) \\]

Thus, our initial equation will rewrite as follows:

\\ [(((5) ^ (2x-3)) \u003d \\ FRAC (1) (25) \\ RIGHTARROW ((5) ^ (2x-3)) \u003d ((5) ^ (- 2)) \\]

But this is already quite solved! On the left in the equation there is an indicative function, the right in the equation is the indicative function, nothing but they are no longer anywhere. Consequently, it is possible to "discard" the foundations and stupidly equate the indicators:

Received the simplest linear equation, which any student will decide literally in a couple of lines. Well, in four lines:

\\ [\\ begin (align) & 2x-3 \u003d -2 \\\\ & 2x \u003d 3-2 \\\\ & 2x \u003d 1 \\\\ & x \u003d \\ FRAC (1) (2) \\\\ End (Align) \\]

If you do not understand what now happened in the last four lines - be sure to return to the topic "Linear equations" and repeat it. Because without a clear assimilation of this topic, it is too early for the indicative equations.

\\ [((9) ^ (x)) \u003d - 3 \\]

Well, how to solve this? The first thought: $ 9 \u003d 3 \\ Cdot 3 \u003d (((3) ^ (2)) $, so the initial equation can be rewritten so:

\\ [((\\ left (((3) ^ (2)) \\ right)) ^ (x)) \u003d - 3 \\]

Then you remember that when the degree is raised into the degree, the indicators are variable:

\\ [((\\ left (((3) ^ (2)) \\ RIGHT)) ^ (x)) \u003d ((3) ^ (2x)) \\ Rightarrow ((3) ^ (2x)) \u003d - (( 3) ^ (1)) \\]

\\ [\\ begin (align) & 2x \u003d -1 \\\\ & x \u003d - \\ FRAC (1) (2) \\\\\\ End (Align) \\]

And here for such a decision we will get honestly deserved two. For we with the calm of the Pokemon sent a "minus" sign, facing the top three, to the degree of this troika. And so do it is impossible. And that's why. Take a look at different degrees of the troika:

\\ [\\ begin (Matrix) ((3) ^ (1)) \u003d 3 ° ((3) ^ (- 1)) \u003d \\ FRAC (1) (3) & ((3) ^ (\\ FRAC (1) ( 2))) \u003d \\ sqrt (3) \\\\ ((3) ^ (2)) \u003d 9 Δ (3) ^ (- 2)) \u003d \\ FRAC (1) (9) & ((3) ^ (\\ 3) ^ (- \\ FRAC (1) (2))) \u003d \\ FRAC (1) (\\ SQRT (3)) \\\\\\ End (Matrix) \\]

By making this sign, I just did not perverted: and considered positive degrees, and negative, and even fractional ... so where is at least one negative number? His not! And can not be because the indicative function is $ y \u003d (((a) ^ (x)) $, first, always takes only positive values \u200b\u200b(how many units do not multiply or not delivered to a twice - there will still be a positive number), And secondly, the basis of such a function is the number $ a $ - by definition is a positive number!

Well, how then to solve the equation $ ((9) ^ (x)) \u003d - $ 3? But in no way: there are no roots. And in this sense, the indicative equations are very similar to square - there may also not be roots. But if in square equations, the number of roots is determined by the discriminant (discriminant positive - 2 roots, negative - no roots), then everything depends on what is worth the right of the equality sign.

Thus, we formulate a key conclusion: the simplest indicative equation of the type $ ((a) ^ (x)) \u003d b $ has a root then and only if $ B \\ gt 0 $. Knowing this simple fact, you can easily determine: there is a root equation proposed for you or not. Those. Is it worth it to solve it or immediately write down that there are no roots.

This knowledge will still repeatedly help us when you have to solve more complex tasks. In the meantime, the lyrics are enough - it's time to study the main algorithm for solving indicative equations.

How to solve exponential equations

So, we formulate the task. It is necessary to solve the indicative equation:

\\ [((a) ^ (x)) \u003d b, \\ quad a, b \\ gt 0 \\]

According to the "naive" algorithm, through which we have earlier, it is necessary to present the number $ b $ as the degree of $ A $:

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

\\ [\\ begin (align) & ((2) ^ (x)) \u003d 8 \\ rightarrow ((2) ^ (x)) \u003d ((2) ^ (3)) \\ rightarrow x \u003d 3; \\\\ & ((3) ^ (- x)) \u003d 81 \\ rightarrow ((3) ^ (- x)) \u003d ((3) ^ (4)) \\ rightarrow -x \u003d 4 \\ rightarrow x \u003d -4; \\\\ & ((5) ^ (2x)) \u003d 125 \\ rightarrow ((5) ^ (2x)) \u003d ((5) ^ (3)) \\ Rightarrow 2x \u003d 3 \\ Rightarrow x \u003d \\ FRAC (3) ( 2). \\\\\\ End (Align) \\]

And oddly enough, this scheme works in about 90% of cases. And then with the rest of 10%? The remaining 10% is a bit "schizophrenic" indicative equations of the form:

\\ [((2) ^ (x)) \u003d 3; \\ quad ((5) ^ (x)) \u003d 15; \\ quad ((4) ^ (2x)) \u003d 11 \\]

Well, what extent you need to build 2 to get 3? First? And here is not: $ ((2) ^ (1)) \u003d 2 $ - not enough. In the second? There is also no: $ ((2) ^ (2)) \u003d $ 4 - a bit too much. And in which then?

Knowing students already probably guessed: in such cases, when "beautifully" cannot be solved, "heavy artillery" - logarithms are connected. Let me remind you that with the help of logarithms, any positive number can be represented as a degree of any other positive number (except for one):

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

\\ [\\ begin (Align) & ((2) ^ (x)) \u003d 3 \\\\ & \u003d (((b) ^ (((\\ log) _ (b)) a)) \\\\\\ End (Align) \\]

If we assume that $ a \u003d $ 3 is our original number, which is worth the right, and $ b \u003d 2 $ is the most base of the indicative function to which we want to bring the right part so that we obtain the following:

\\ [\\ begin (align) & a \u003d ((b) ^ (((\\ log) _ (b)) a)) \\ rightarrow 3 \u003d (((2) ^ (((\\ log) _ (2)) 3 )); \\\\ & ((2) ^ (x)) \u003d 3 \\ rightarrow ((2) ^ (x)) \u003d (((2) ^ (((\\ log) _ (2)) 3)) \\ Rightarrow x \u003d ( (\\ log) _ (2)) 3. \\\\\\ End (Align) \\]

Received a little strange answer: $ x \u003d ((\\ log) _ (2)) $ 3. In some other task, many would be laughed in such an answer and began to recheck their solution: suddenly there was a mistake somewhere? I hurry to refress you: no error is not here, and logarithm in the roots of the indicative equations is a completely typical situation. So get used to. :)

Now we decide by the analogy of the remaining two equations:

\\ [\\ begin (align) & ((5) ^ (x)) \u003d 15 \\ rightarrow ((5) ^ (x)) \u003d ((5) ^ (((\\ log) _ (5)) 15)) \\ Rightarrow x \u003d ((\\ log) _ (5)) 15; \\\\ Δ ((4) ^ (2x)) \u003d 11 \\ rightarrow ((4) ^ (2x)) \u003d ((4) ^ (((\\ log) _ (4)) 11)) \\ Rightarrow 2x \u003d ( (\\ log) _ (4)) 11 \\ RIGHTARROW X \u003d \\ FRAC (1) (2) ((\\ Log) _ (4)) 11. \\\\\\ End (Align) \\]

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

This we made a multiplier to the argument of Logarithm. But no one prevents us from making this multiplier to the ground:

In this case, all three options are correct - these are simply different forms of recording of the same number. Which one to choose and write down in the present decision - to solve only you.

Thus, we learned how to solve any indicative equations of the type $ ((a) ^ (x)) \u003d b $, where the numbers $ a $ and $ b $ are strictly positive. However, the harsh reality of our world is such that such simple tasks will meet you very and very rarely. Much more often you will come across something like this:

\\ [\\ begin (align) & ((4) ^ (x)) + ((4) ^ (x - 1)) \u003d ((4) ^ (x + 1)) - 11; \\\\ & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & ((100) ^ (x - 1)) \\ Cdot ((2.7) ^ (1-x)) \u003d 0.09. \\\\\\ End (Align) \\]

Well, how to solve this? Is it possible to solve? And if so, how?

Without panic. All these equations quickly and simply reduce the simple formulas that we have already considered. Just need to know remember a couple of techniques from the course of algebra. And of course, here is nowhere without rules for working with degrees. About this I will tell you now. :)

Transformation of indicative equations

The first thing to remember is: any indicative equation, no matter how difficult it is, anyway, should be reduced to the simplest equations - thereby we have already considered and which we know how to solve. In other words, the scheme of solving any indicative equation is as follows:

1. Record the source equation. For example: $ ((4) ^ (x)) + ((4) ^ (x-1)) \u003d ((4) ^ (x + 1)) - 11 $;
2. Make some incomprehensible crap. Or even a few horses, which are called "convert equation";
3. At the output to obtain the simplest expressions of the type $ ((4) ^ (x)) \u003d $ 4 or something else in this spirit. Moreover, one initial equation can give several such expressions at once.

With the first item, everything is clear - even my cat will be able to record the equation on the leaf. With the third point, too, it seems, more or less clearly - we have already groaned such equations.

But how to be with the second item? What kind of transformation? What to convert in? And How?

Well, let's understand. First of all, I will note the following. All indicative equations are divided into two types:

1. The equation is composed of indicative functions with the same base. Example: $ ((4) ^ (x)) + ((4) ^ (x - 1)) \u003d ((4) ^ (x + 1)) - 11 $;
2. The formula has demonstration functions with different bases. Examples: $ ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)) $ and $ ((100) ^ (x-1) ) \\ Cdot ((2.7) ^ (1-x)) \u003d 0.09 $.

Let's start with the equations of the first type - they are solved the easiest. And in their solution, we will help such a reception as the allocation of sustainable expressions.

Allocation of a stable expression

Let's look at this equation again:

\\ [((4) ^ (x)) + ((4) ^ (x - 1)) \u003d ((4) ^ (x + 1)) - 11 \\]

What do we see? The fourthkee is erected in various degrees. But all these degrees are the simple amounts of the $ x $ variable with other numbers. Therefore, it is necessary to recall the rules for working with degrees:

\\ [\\ begin (align) & ((a) ^ (x + y)) \u003d ((a) ^ (x)) \\ Cdot ((a) ^ (y)); \\\\ & ((a) ^ (xy)) \u003d ((a) ^ (x)): ((a) ^ (y)) \u003d \\ FRAC (((a) ^ (x))) ((( ) ^ (y))). \\\\\\ End (Align) \\]

Simply put, the addition of indicators can be converted into the work of degrees, and the subtraction is easily converted into division. Let's try to apply these formulas to degrees from our equation:

\\ [\\ Begin (Align) & ((4) ^ (x - 1)) \u003d \\ FRAC (((4) ^ (x))) (((4) ^ (1))) \u003d ((4) ^ (x)) \\ Cdot \\ FRAC (1) (4); \\\\ · ((4) ^ (x + 1)) \u003d ((4) ^ (x)) \\ Cdot ((4) ^ (1)) \u003d ((4) ^ (x)) \\ CDOT 4. \\ I rewrite the original equation, taking into account this fact, and then collect all the components on the left:

\\ [\\ begin (align) & ((4) ^ (x)) + ((4) ^ (x)) \\ Cdot \\ FRAC (1) (4) \u003d ((4) ^ (x)) \\ CDOT 4 -eleven; \\\\ & ((4) ^ (x)) + ((4) ^ (x)) \\ Cdot \\ FRAC (1) (4) - ((4) ^ (x)) \\ Cdot 4 + 11 \u003d 0. \\\\\\ End (Align) \\]

In the first four components there is an element $ ((4) ^ (x)) $ - I will bring it for the bracket:

\\ [\\ begin (align) & ((4) ^ (x)) \\ cdot \\ left (1+ \\ FRAC (1) (4) -4 \\ RIGHT) + 11 \u003d 0; \\\\ & ((4) ^ (x)) \\ CDOT \\ FRAC (4 + 1-16) (4) + 11 \u003d 0; \\\\ & ((4) ^ (x)) \\ Cdot \\ left (- \\ FRAC (11) (4) \\ Right) \u003d - 11. \\\\\\ End (Align) \\]

It remains to divide both parts of the equation for the fraction of $ - \\ FRAC (11) (4) $, i.e. Essentially multiply to the overtook fraction - $ - \\ FRAC (4) (11) $. We get:

\\ [\\ Begin (Align) & ((4) ^ (x)) \\ Cdot \\ left (- \\ FRAC (11) (4) \\ RIGHT) \\ CDOT \\ LEFT (- \\ FRAC (4) (11) \\ RIGHT ) \u003d - 11 \\ Cdot \\ left (- \\ FRAC (4) (11) \\ RIGHT); \\\\ & ((4) ^ (x)) \u003d 4; \\\\ & ((4) ^ (x)) \u003d ((4) ^ (1)); \\\\ & x \u003d 1. \\\\\\ End (Align) \\]

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

At the same time, in the process of solutions, we found (and even carried out for the bracket) the total multiplier $ ((4) ^ (x)) $ is a stable expression. It can be denoted by a new variable, and you can simply gently express and get the answer. In any case, the key principle of solving the following:

Find a stable expression in the source equation containing a variable that is easily highlighted from all the indicative functions.

The good news is that almost every indicative equation allows the allocation of such a stable expression.

But there are bad news: such expressions may be very cunning, and it is quite difficult to allocate them. Therefore, we will analyze another task:

\\ [((5) ^ (x + 2)) + ((0.2) ^ (- x - 1)) + 4 \\ Cdot ((5) ^ (x + 1)) \u003d 2 \\]

Perhaps someone will now have a question: "Pasha, what did you whistle? Here, different bases - 5 and 0.2 ". But let's try to convert a degree with a base of 0.2. For example, get rid of decimal fractions, bringing it to normal:

\\ [((0.2) ^ (- x - 1)) \u003d (((0.2) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (2) (10 ) \\ Right)) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (- \\ left (x + 1 \\ right)) ) \\]

As you can see, the number 5 after all appeared, let it both in the denominator. At the same time rewrote the indicator in the form of a negative. And now I remember one of the most important rules for working with degrees:

\\ [((a) ^ (- n)) \u003d \\ FRAC (1) (((a) ^ (n))) \\ Rightarrow ((\\ left (\\ FRAC (1) (5) \\ Right)) ^ ( - \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (5) (1) \\ RIGHT)) ^ (x + 1)) \u003d ((5) ^ (x + 1)) \\ Here I, of course, slightly rushed. Because for a complete understanding of the formula of deliverance from negative indicators, it was necessary to record like this:

\\ [((a) ^ (- n)) \u003d \\ FRAC (1) (((a) ^ (n))) \u003d ((\\ left (\\ FRAC (1) (A) \\ RIGHT)) ^ (n )) \\ Rightarrow ((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (\\ FRAC (5) (1) \\ On the other hand, nothing prevented us to work with one shot:

\\ [((\\ left (\\ FRAC (1) (5) \\ RIGHT)) ^ (- \\ left (x + 1 \\ right))) \u003d ((\\ left (((5) ^ (- 1)) \\ )) \u003d ((5) ^ (x + 1)) \\]

But in this case, you need to be able to erect a degree to another degree (remind you: the indicators are folded). But I did not have to "turn over" the fractions - perhaps for someone it will be easier. :)

In any case, the initial indicative equation will be rewritten as:

\\ [\\ begin (align) & ((5) ^ (x + 2)) + ((5) ^ (x + 1)) + 4 \\ cdot ((5) ^ (x + 1)) \u003d 2; \\\\ & ((5) ^ (x + 2)) + 5 \\ Cdot ((5) ^ (x + 1)) \u003d 2; \\\\ & ((5) ^ (x + 2)) + ((5) ^ (1)) \\ Cdot ((5) ^ (x + 1)) \u003d 2; \\\\ & ((5) ^ (x + 2)) + ((5) ^ (x + 2)) \u003d 2; \\\\ & 2 \\ Cdot ((5) ^ (x + 2)) \u003d 2; \\\\ & ((5) ^ (x + 2)) \u003d 1. \\\\\\ End (Align) \\]

So it turns out that the initial equation is even easier than the previously considered: there is no need to allocate a steady expression - everything itself has decreased. It remains only to recall that $ 1 \u003d ((5) ^ (0)) $, from where we get:

\\ [\\ begin (align) & ((5) ^ (x + 2)) \u003d ((5) ^ (0)); \\\\ & x + 2 \u003d 0; \\\\ & x \u003d -2. \\\\\\ End (Align) \\]

That's all the decision! We got the final answer: $ x \u003d -2 $. At the same time I would like to note one reception, which greatly simplified us all the calculations:

In the lower equations, be sure to get rid of decimal fractions, translate them into ordinary. This will allow you to see the same foundations of degrees and will significantly simplify the decision.

Let us now turn to more complex equations in which there are different foundations that are not at all reduced to each other with the help of degrees.

Use the properties of degrees

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

\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & ((100) ^ (x - 1)) \\ Cdot ((2.7) ^ (1-x)) \u003d 0.09. \\\\\\ End (Align) \\]

The main difficulty here is not clear what to bring to what basis. Where are the stable expressions? Where are the same foundations? There is no need for it.

But let's try to go to another way. If there are no ready-made values, you can try to find, laying out the reasons for multipliers.

Let's start with the first equation:

\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & 21 \u003d 7 \\ CDOT 3 \\ RIGHTARROW ((21) ^ (3X)) \u003d ((\\ left (7 \\ CDOT 3 \\ RIGHT)) ^ (3x)) \u003d ((7) ^ (3x)) \\ \\\\\\ End (Align) \\]

But after all, you can proceed on the contrary - make up from numbers 7 and 3 number 21. Especially it is easy to do on the left, since the indicators and both degrees are the same:

\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ Cdot ((3) ^ (x + 6)) \u003d ((\\ left (7 \\ CDOT 3 \\ RIGHT)) ^ (x + 6)) \u003d ((21) ^ (x + 6)); \\\\ & ((21) ^ (x + 6)) \u003d ((21) ^ (3x)); \\\\ & x + 6 \u003d 3x; \\\\ & 2x \u003d 6; \\\\ & x \u003d 3. \\\\\\ End (Align) \\]

That's all! You made an indicator of the degree outside the work and immediately got a beautiful equation, which is solved in a couple of lines.

Now we will deal with the second equation. Everything is much more difficult here:

\\ [((100) ^ (x - 1)) \\ CDOT ((2.7) ^ (1-x)) \u003d 0.09 \\]

\\ [((100) ^ (x - 1)) \\ CDOT ((\\ left (\\ FRAC (27) (10) \\ RIGHT)) ^ (1-x)) \u003d \\ FRAC (9) (100) \\]

In this case, the fractions were disracotized, but if something could be reduced - be sure to reduce. Often, at the same time, interesting grounds will appear with which you can already work.

Also, unfortunately, nothing really appeared. But we see that the indicators of degrees standing in the work on the left are opposite:

Let me remind you: to get rid of the "minus" sign in the indicator, it is enough to "turn over" the fraction. Well, rewrite the original equation:

\\ [\\ begin (align) & ((100) ^ (x - 1)) \\ CDOT ((\\ left (\\ FRAC (10) (27) \\ RIGHT)) ^ (x - 1)) \u003d \\ FRAC (9 )(100); \\\\ \\ (\\ left (100 \\ CDOT \\ FRAC (10) (27) \\ RIGHT)) ^ (x - 1)) \u003d \\ FRAC (9) (100); \\\\ & ((\\ left (\\ FRAC (1000) (27) \\ RIGHT)) ^ (X - 1)) \u003d \\ FRAC (9) (100). \\\\\\ End (Align) \\]

In the second line, we simply carried out a general figure from the work for a bracket according to the rule $ ((a) ^ (x)) \\ Cdot ((b) ^ (x)) \u003d ((\\ left (a \\ cdot b \\ right)) ^ (x)) $, and in the latter just multiplied the number 100 by fraction.

Now we note that the numbers standing on the left (at the base) and on the right, are alike. Than? Yes, obviously: they are degrees of the same number! We have:

\\ [\\ begin (align) \\ FRAC (1000) (27) \u003d \\ FRAC (((10) ^ (3))) (((3) ^ (3))) \u003d ((\\ left (\\ FRAC ( 10) (3) \\ RIGHT)) ^ (3)); \\\\ \\ FRAC (9) (100) \u003d \\ FRAC (((3) ^ (2))) (((10) ^ (3))) \u003d ((\\ left (\\ FRAC (3) (10) \\ Right)) ^ (2)). \\\\\\ End (Align) \\]

Thus, our equation will rewrite as follows:

\\ [((\\ left ((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3)) \\ Right)) ^ (x - 1)) \u003d ((\\ left (\\ FRAC (3 ) (10) \\ RIGHT)) ^ (2)) \\]

\\ [((\\ left (((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3)) \\ Right)) ^ (x - 1)) \u003d ((\\ left (\\ FRAC (10 ) (3) \\ Right)) ^ (3 \\ left (x - 1 \\ right))) \u003d ((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3x-3)) \\]

At the same time, you can also get a degree with the same basis, for which it is enough to "turn over" the fraction:

\\ [((\\ left (\\ FRAC (3) (10) \\ RIGHT)) ^ (2)) \u003d ((\\ left (\\ FRAC (10) (3) \\ Right)) ^ (- 2)) \\]

Finally, our equation will take the form:

\\ [\\ begin (Align) & ((\\ Left (\\ FRAC (10) (3) \\ RIGHT)) ^ (3x-3)) \u003d ((\\ left (\\ FRAC (10) (3) \\ RIGHT)) ^ (- 2)); \\\\ & 3x-3 \u003d -2; \\\\ & 3x \u003d 1; \\\\ & X \u003d \\ FRAC (1) (3). \\\\\\ End (Align) \\]

That's the whole decision. His main idea is reduced to the fact that even under different reasons, we are trying by any truths and inconsistencies to reduce these grounds for the same. This is helped by elementary transformations of equations and rules for working with degrees.

But what are the rules and when to use? How to understand that in one equation you need to share both sides for something, and in the other - to lay the basis of the indicative function on multipliers?

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

And to help you in this hard matter, I propose to download a set of equations for an independent solution on my site. To all equations there are answers, so you can always check yourself.

In general, I wish you a good workout. And you will see you in the next lesson - there we will disassemble the truly complex lower equations, where the methods described above are not enough. And the simple training will also be not enough. :)

The solution of most mathematical problems is one way or another associated with the transformation of numerical, algebraic or functional expressions. Specified in particular applies to the decision. In the variants of the EGE in mathematics to such a type of tasks, the task is, in particular, the C3 problem. Learning to solve the tasks C3 It is important not only for the purpose of successful delivery of the USE, but for the reason that this skill is useful when studying the course of mathematics in the highest school.

Performing tasks C3 has to solve various types of equations and inequalities. Among them are rational, irrational, indicative, logarithmic, trigonometrics containing modules (absolute values), as well as combined. This article discusses the main types of indicative equations and inequalities, as well as various methods of their solutions. Read about the decision of the remaining types of equations and inequalities in the heading "" in the articles on the methods of solving the problems of C3 from the OPER options in mathematics.

Before proceeding with the analysis of specific indicative equations and inequalitiesAs a tutor in mathematics, I suggest you refreshing some theoretical material that we need.

Exponential function

What is an indicative function?

Function of type y. = a X.where a. \u003e 0 I. a. ≠ 1, called indicative function.

Maintenance properties of the indicative function y. = a X.:

Graph indicative function

The graph of the indicative function is exhibitor:

Graphs of indicative functions (exhibitors)

Solution of indicative equations

Indicative They are called equations in which an unknown variable is only in the indicators of any degrees.

For solutions indicatory equations You need to know and be able to use the following simple theorem:

Theorem 1. Indicative equation a. f.(x.) = a. g.(x.) (where a. > 0, a. ≠ 1) equivalent equation f.(x.) = g.(x.).

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

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Example 1. Solve the equation:

Decision: We use the above formulas and substitution:

The equation then takes the form:

The discriminant of the obtained square equation is positive:

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

Turning to the return substitution, we get:

The second root equation does not have, since the indicative function is strictly positive throughout the entire definition area. We solve the second:

Taking into account the said in Theorem 1, go to the equivalent equation: x. \u003d 3. This will be the answer to the task.

Answer: x. = 3.

Example 2. Solve the equation:

Decision: There are no restrictions on the area of \u200b\u200bpermissible values \u200b\u200bat the equation, since the feeding expression makes sense at any meaning x. (exponential function y. = 9 4 -X. positive and not equal to zero).

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

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

Answer:x.= 6.

Example 3. Solve the equation:

Decision: Both parts of the source equation can be divided by 0.2 x. . This transition will be equivalent, since this expression is greater than zero in any value. x. (The indicative function is strictly positive on its definition area). Then the equation takes the form:

Answer: x. = 0.

Example 4. Solve the equation:

Decision:we simplify the equation to elementary by equivalent transformations using the rules of division and multiplication of degrees given at the beginning of the article:

Division of both parts of the equation for 4 x. , as in the previous example, is an equivalent transformation, since this expression is not zero at no matter what values x..

Answer: x. = 0.

Example 5. Solve the equation:

Decision: function y. = 3 X., standing in the left side of the equation, is increasing. Function y. = —x.-2/3, standing in the right part of the equation, is descending. This means that if the graphs of these functions intersect, then not more than one point. In this case, it is not difficult to guess that the graphs intersect at the point x. \u003d -1. There will be no other roots.

Answer: x. = -1.

Example 6. Solve the equation:

Decision: We simplify the equation by equivalent transformations, bearing in mind everywhere that the indicative function is strictly greater than zero in any meaning x.and using the rules for calculating the work and private degrees given at the beginning of the article:

Answer: x. = 2.

Solution of indicative inequalities

Indicative It is called inequalities in which an unknown variable is contained only in the indicators of any degrees.

For solutions indicative inequalities Knowledge requires the following theorem:

Theorem 2. If a a. \u003e 1, then inequality a. f.(x.) > a. g.(x.) It 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.) It is equivalent to the inequality of the opposite sense: f.(x.) < g.(x.).

Example 7.Solve inequality:

Decision: Imagine initial inequality in the form:

We divide both parts of this inequality for 3 2 x. at the same time (due to the positivity of the function y.= 3 2x.) The sign of inequality will not change:

We use the substitution:

Then the inequality will take the form:

So, the solution of inequality is the gap:

turning to the return substitution, we get:

The left inequality due to the positiveness of the indicative function is performed automatically. Taking advantage of the well-known property of the logarithm, proceed to the equivalent inequality:

Since there will be a transition to the following inequality to the following in the degree of degree, there will be a transition to the following inequality:

So, finally get answer:

Example 8. Solve inequality:

Decision: Using the properties of multiplication and dividing degrees, rewrite inequality in the form:

We introduce a new variable:

Taking into account this substitution, inequality takes the form:

Multiply the numerator and denominator of the fraction on 7, we get the following equivalent inequality:

So, inequality satisfy the following values \u200b\u200bof the variable t.:

Then, passing to the return substitution, we get:

Since the foundation of the degree here is more than a unit, equivalent (by Theorem 2) will transition to inequality:

Finally get answer:

Example 9. Solve inequality:

Decision:

We divide both parts of inequality to the expression:

It is always greater than zero (due to the positiveness of the indicative function), so the sign of inequality is not necessary. We get:

t, located in the interval:

Turning to the return substitution we obtain that the initial inequality disintegrates into two cases:

The first inequality of solutions does not have the relevance of the indicative function. We solve the second:

Example 10. Solve inequality:

Decision:

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

Parabola branches y. = x. 2 -2x.+2, standing in the indicator, are directed up, it means it is limited to the bottom with the value that it reaches in its vertex:

Together with this limited bottom, the function is also y. = 3 x. 2 -2x.+2, standing in the right part of the equation. It reaches its smallest value at the same point as the parabola standing in the indicator, and this value is 3 1 \u003d 3. So, the initial inequality may be correct only if the function on the left and the function on the right is taken at one point the value equal to 3 (the intersection of areas of these functions is only this number). This condition is performed in a single point. x. = 1.

Answer: x.= 1.

In order to learn to decide indicative equations and inequalities, It is necessary to constantly train in their decision. In this difficult case, various methodological manuals can help you, collectors for elementary mathematics, collections of competitive tasks, classes in mathematics at school, as well as individual classes with a professional tutor. I sincerely wish you success in the preparation and brilliant results on the exam.

Sergey Valerievich

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

At the stage of preparation for final testing of high school students, you need to tighten the knowledge on the topic "Indective Equations". The experience of past years indicates that such tasks cause certain difficulties from schoolchildren. Therefore, high school students, regardless of their level of preparation, it is necessary to carefully assimilate the theory, remember the formulas and understand the principle of solving such equations. Having managed to cope with this type of tasks, graduates will be able to count on high points when passing the exam in mathematics.

Get ready for exam testing with "Shkolkovo"!

When repeating the materials passed, many students face the problem of finding the formulas necessary to solve the equations. The school textbook is not always at hand, and the selection of the necessary information on the topic on the Internet takes a long time.

The educational portal "Skolkovo" offers students to take advantage of our knowledge base. We implement a completely new method of preparation for final testing. When doing on our website, you can identify gaps in knowledge and pay attention to those assignments that cause the greatest difficulties.

Teachers "Shkolkovo" collected, systematized and outlined all the material necessary for the successful passing of the EGE as simple as possible and accessible form.

The main definitions and formulas are presented in the "Theoretical Help" section.

For better assimilation of the material, we recommend practicing the tasks. Carefully view examples of exponential equations on this page to understand the calculation algorithm. After that, proceed to perform tasks in the "Catalogs" section. You can start with the easiest tasks or immediately move to solving complex indicative equations with several unknown or. The exercise base on our site is constantly complemented and updated.

Those examples with the indicators that have trouble make it possible to add to favorites. So you can quickly find them and discuss the decision with the teacher.

To successfully pass the exam, engage in the "Shkolkovo" portal every day!

What is the indicative equation? Examples.

So, the indicative equation ... A new unique exhibit at our overall exhibition of a wide variety of equations!) How it almost always happens, the key word of any new mathematical term is the corresponding adjective, which is characterized by it. So here. The key word in the term "indicative equation" is the word "Indective". What does it mean? This word means that unknown (X) is in the indicators of any degrees. And only there! It is extremely important.

For example, such simple equations:

3 x +1 \u003d 81

5 x + 5 x +2 \u003d 130

4 · 2 2 x -17 · 2 x +4 \u003d 0

Or even such monsters:

2 SIN x \u003d 0.5

Please immediately pay attention to one important thing: in basins degrees (bottom) - only numbers. But B. indicators degrees (from above) - a wide variety of expressions with Xa. Completely any.) Everything depends on the specific equation. If, suddenly, the EX will come out in the equation anywhere else, in addition to the indicator (say, 3 x \u003d 18 + x 2), then such an equation will be already equation mixed type. Such equations do not have clear rules for solutions. Therefore, in this lesson we will not consider them. To the joy of disciples.) Here we will consider only the indicative equations in the "clean" form.

Generally speaking, even clean indicative equations are clearly solved far from everything and not always. But among the whole rich diversity of exponential equations there are certain types that can be solved and needed. That's these types of equations we will look at. And the examples are definitely shaking.) So you are comfortable and - on the road! As in the computer "shooting", our journey will be held in levels.) From elementary to simple, from simple - to the middle and middle - to complex. On the way you will also be waiting for the secret level - the receptions and methods for solving non-standard examples. Those who you will not read in most school textbooks ... Well, at the end you, of course, waiting for the final boss in the form of the houses.)

Level 0. What is the simplest indication equation? The solution of the simplest indicative equations.

To begin with, consider some frank elementary. From something you need to start, right? For example, such an equation:

2 x \u003d 2 2

Even without any theories, on simple logic and common sense, it is clear that x \u003d 2. otherwise it is not true? No other value of the ICA is good ... And now we will turn our eyes on record decision of this cool indicative equation:

2 x \u003d 2 2

X \u003d 2.

What happened to us? And the following happened. We actually took and ... just thrown out the same bases (twos)! They completely thrown away. And what pleases, got into the apple!

Yes, indeed, if in the exponential equation to the left and right the samenumbers in any degrees, then these numbers can be discarded and simply equated degrees. Mathematics permits.) And then you can work separately with the indicators and decide a much simpler equation. Great, right?

Here is the key idea of \u200b\u200bsolving any (yes, it is anyone!) Indicative equation: with the help of identical transformations, it is necessary to ensure that the left and right in the equation stood the same Round numbers in various degrees. And then you can safely remove the same bases and equate the indicators of degrees. And work with a simpler equation.

Now I remember the Iron Rule: it is possible to remove the same bases if and only if in the equation on the left and right, the grounds are worth in proud loneliness.

What does it mean in proud loneliness? This means without any neighbors and coefficients. I explain.

For example, in the equation

3 · 3 x-5 \u003d 3 2 x +1

Troika can not be removed! Why? Because on the left we have not just a lonely three to the degree, but composition 3 · 3 X-5. Excess Troika interferes: the coefficient, you understand.)

The same can be said about the equation

5 3 x \u003d 5 2 x +5 x

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

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

a. F. ( X.) = a G. ( X.)

This type of indicative equation is called simplest. Or, scientifically, canonical . And whatever the raised equation before us is, we will have it, anyway, we will reduce exactly the simplest (canonomic) mind. Or, in some cases, to total equations of this type. Then our simplest equation can generally rewrite like this:

F (x) \u003d g (x)

And that's it. This will be equivalent to the transformation. In this case, as f (x) and G (x), completely any expressions with the X can stand. Any please.

Perhaps a particularly inquisitive student will ask: And from what basis are we so easily and simply discarding the same bases on the left and right and equate the indicators of degrees? Intuition intuition, but suddenly, in some equation and for some reason this approach will be incorrect? Does it always legally throw out the same foundations? Unfortunately, for a strict mathematical answer, this interesting question is needed quite deeply and seriously immersed in the general theory of the device and behavior of functions. A little more specifically - in the phenomenon strict monotony. In particular, strict monotony Indicative functiony.= a X.. Since it is the indicative function and its properties that underlie the solution of the indicative equations, yes.) The deployed response to this question will be given in a separate special system dedicated to solving complex non-standard equations using the monotony of different functions.)

To explain in detail this moment now is just to bring the brain to the average student and scareten it ahead of time with dry and charming theory. I will not do this.) For our main task is at the moment - learn to solve the demonstration equations! The most simplest! Therefore, they don't care and boldly throw out the same foundations. it can, Believe me for the word!) And then already solve the equivalent equation f (x) \u003d g (x). As a rule, simpler than the original is indicative.

Of course, it is assumed that it is already able to solve even the equations without ICs in the indicators, the people are already able at the moment.) Who still does not know how - boldly closing this page, walk along the relevant references and replenish old gaps. Otherwise, you will have to you have anything, yes ...

I'm silent about irrational, trigonometric and other brutal equations, which can also emerge in the process of eliminating grounds. But do not be afraid, frank tin in the indicators of degrees, we will not consider it: it's too early. We will train only on the most ordinary equations.)

Now consider equations that require some additional efforts to give them to the simplest. For differences we will call them simple indicative equations. So, moving to the next level!

Level 1. Simple demonstration equations. Recognize degree! Natural indicators.

Key rules in solving any indicative equations are the rules of action with degrees. Without these knowledge and skills, nothing will happen. Alas. So, if with degrees of the problem, then I ask for the beginning of the grace. In addition, we need us. These transformations (as many as two!) - the basis of the solution of all mathematics equations in general. And not only indicative. So, who forgot, also stroll through the reference: I don't just put them on them.

But one only actions with degrees and identical transformations are few. Still needed personal observation and an increase. We are needed the same foundations, right? So we look at the example and we are looking for them in an apparent or disguised form!

For example, such an equation:

3 2 x - 27 x +2 \u003d 0

First look by basis. They are different! Troika and twenty seven. But to panic and fall into despair early. It's time to remember that

27 = 3 3

Numbers 3 and 27 - relative to the degree! And close.) It became, we have the full right to write:

27 x +2 \u003d (3 3) x + 2

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

(a m) n \u003d a Mn

If you continue to run it, then it is generally perfect:

27 x +2 \u003d (3 3) x + 2 \u003d 3 3 (x +2)

The initial example now looks like this:

3 2 x - 3 3 (x +2) \u003d 0

Excellent, the foundations of the degrees were leveled. What we have sought. Frequently done.) But now we run a basic identity conversion into the course - tolerate 3 3 (x +2) to the right. Nobody canceled the elementary actions of mathematics, yes.) We get:

3 2 x \u003d 3 3 (x +2)

What does this type of equation give us? And the fact that now our equation is reduced to canonical appearance: On the left and right are the same numbers (Troika) in degrees. And both troops are in proud loneliness. We boldly remove the troika and get:

2x \u003d 3 (x + 2)

We decide this and get:

X \u003d -6.

That's all things. This is the correct answer.)

And now comprehend the decision of the solution. What was saved in this example? We were saved by the knowledge of the Troika degrees. How exactly? we identified Among the 27 encrypted troika! This receiver (encryption of the same base under different numbers) is one of the most popular in the indicative equations! Unless the most popular. Yes, and also, by the way. That is why in the indicative equations, the observation and ability to recognize in numbers of other numbers are so important!

Practical advice:

The degrees of popular numbers need to know. In face!

Of course, we build a two-seventh degree or three in the fifth. Not in the mind, so at least in the draft. But in the demonstration equations, it is much more often necessary to be degree, but on the contrary - to learn what number and to what extent is hidden at the number, say, 128 or 243. And this is already more complicated than simple construction, you will agree. Feel the difference what is called!

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

To determine what degrees 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 (corrosion, 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 demonstration equations. Recognize degree! Negative and fractional indicators.

At this level, we already use our knowledge of degrees to the full coil. Namely - involve negative and fractional indicators in this fascinating process! Yes Yes! We need to build power, right?

For example, such a terrible equation:

Again, first glance is on the ground. Basins are different! Moreover, this time is even remotely not similar to each other! 5 and 0.04 ... And to eliminate grounds, you need the same ... What to do?

Nothing wrong! In fact, everything is the same, just a bond between the top and 0.04 visually visible badly. How to get out? And let's turn around 0.04 to the usual fraction! And there, you look, everything is formed.)

0,04 = 4/100 = 1/25

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

How? Now the connection between numbers 5 and 1/25 is easier to coal? That's what it is ...

And now according to the rules of action with degrees with negative indicatoryou can write a solid hand:

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

Again, according to the rules of action with degrees, you can now write:

(5 -2) x -1 \u003d 5 -2 (x -1)

Just in case, I remind you (suddenly, who is not aware) that the basic rules of action with degrees are fair for any Indicators! Including for negative.) So boldly take and replacing the indicators (-2) and (x - 1) according to the relevant rule. Our equation is getting better and better:

Everything! In addition to lonely tops in the degrees on the left and right there is nothing more. The equation is reduced to canonical form. And then - along the rolling rut. We remove the tops and equate the indicators:

x. 2 –6 x.+5=-2(x.-1)

An example is practically solved. The elementary mathematics of middle classes remained - reveal (correctly!) Braces and collect everything on the left:

x. 2 –6 x.+5 = -2 x.+2

x. 2 –4 x.+3 = 0

We decide this and get two roots:

x. 1 = 1; x. 2 = 3

That's all.)

And now they reflect again. In this example, we again had to recognize the same number in different degrees! Namely - see 0.04 encrypted five. And this time - in negative degree!How did we manage it? From the go - in no way. But after the transition from the decimal fraction 0.04 to the ordinary fraction 1/25 everything and highlighted! And then the whole decision went like oil.)

Therefore, the next green practical council.

If there is decimal fractions in the indicative equation, then we move from decimal fractions to ordinary. In ordinary fractions, it is much easier to recognize the degrees of many popular numbers! After recognition, we move from fractions to degrees with negative indicators.

Keep in mind that such a fint in the indicative equations is found very and very often! And man is not in the subject. It looks, for example, in numbers 32 and 0.125 and is upset. It is not possible to him that this is the same twice, only in different degrees ... But you are already in the subject!)

Solve equation:

In! In appearance - a quiet horror ... However, the appearance of deceptive. This is the simplest indicative equation, despite its awesome appearance. And now I'll show you this.)

First, we are dealt with all the seats sitting in the grounds and in the coefficients. They are clear, different, yes. But we will still risk and try to make them the same! Let's try to get to same number in different degrees. Moreover, it is desirable, the numbers of the most likely small. So, begin to decrypt!

Well, with the fourth, everything is clearly clear - this is 2 2. So, already something.)

With a fraction of 0.25 - it is not clear yet. Need to check. We use the Practical Council - go from a decimal fraction to an ordinary:

0,25 = 25/100 = 1/4

Already much better. For it is now clearly seen that 1/4 is 2 -2. Excellent, and the number 0.25 also freshed with a twos.)

While everything goes well. But the most bad number of all - square root of two! And with this pepper what to do? Is it possible to imagine it as a degree of twos? And who knows him ...

Well, again we climb into our treasury knowledge about degrees! This time you will additionally connect our knowledge about roots. From the course of the 9th grade, we had to bear that any root, if desired, can always be turned into a degree with fractional indicator.

Like this:

In our case:

In how! It turns out, the root square of two is 2 1/2. That's it!

That's fine! All our uncomfortable numbers actually turned out to be encrypted twice.) I do not argue, somewhere very sophisticatedly encrypted. But we, too, also increase their professionalism in the rays of such ciphers! And then everything is obvious. We replace in our equation of 4, 0.25 and the root of two to the degree of two:

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

a m ·a N. = a M. + N.

a M: A N \u003d A M-N

(a m) n \u003d a Mn

For the left side will turn out:

2 -2 · (2 \u200b\u200b2) 5 x -16 \u003d 2 -2 + 2 (5 x -16)

For the right side will be:

And now our evil equation began to look like this:

Who did not handed how this equation was turned out, then the question is not the indicative equations. The question is to action with degrees. I asked urgently to repeat those who have a problem!

Here is the finish line! A canonical view of the indicative equation is obtained! How? I convinced you that was not so scary? ;) We remove the twos and equate the indicators:

It remains only to solve this linear equation. How? With the help of identical transformations, Vestimo.) Dore, what is there! Multiply both parts for a twice (to remove the fraction 3/2), transfer the components with the ICS to the left, without ICS to the right, bring similar, consider - and you will be happy!

Everything should turn out beautifully:

X \u003d 4.

And now again comprehend the course of the solution. In this example, we saw the transition from square root to degrees with an indicator 1/2. And only such a tricky transformation helped us everywhere to go to the same base (two), which saved the position! And if it were not for it, then we would have all the chances to hang out forever and so do not cope with this example, yes ...

Therefore, do not neglect the next practical advice:

If the roots are present in the indictment equation, then go from the roots to degrees with fractional indicators. Very often only such a transformation and clarifies the further situation.

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

It is clear that it is directly erected, for example, a deuce to the degree -3 or the fourth to -3/2 degree is not such a big problem. For knowledgeable.)

But also like, for example, with the move form that

0,125 = 2 -3

Or

Here only practice and rich experience rolls, yes. And, of course, a clear idea, what is negative and fractional degree. And also - practical advice! Yes yes, the very green .) I hope that they will still help you better navigate in the whole diverse diversity of degrees and will significantly increase your chances of success! So do not neglect them. I am not in vain green color I write sometimes.)

But if you become on "you", even with such exotic degrees, as negative and fractional, then your capabilities in solving the explicit equations are volantly expand, and you will already be at the shoulder almost any type of indicative equations. Well, if not any, then interest 80 of all indicative equations - certainly! Yes, yes, I'm not kidding!

So, our first part of the acquaintance with the indicative equations approached its logical conclusion. And, as an intermediate workout, I traditionally offer a little ponslast yourself.)

Exercise 1.

So that my words about deciphering negative and fractional degrees do not disappear, I propose to play a small game!

Imagine the degree of two numbers:

Answers (in disorder):

Happened? Excellent! Then make a combat task - we solve the simplest and simple indicative equations!

Task 2.

Solve equations (all the answers - in disorder!):

5 2x-8 \u003d 25

2 5x-4 - 16 x + 3 \u003d 0

Answers:

x \u003d 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 \u003d 0.5 x · 4 x -4

35 1 - x \u003d 0.2 - x · 7 x

Answers:

x. 1 = -2; x. 2 = 2

x. = 0,5

x. 1 = 3; x. 2 = 5

And these examples of one left? Excellent! You grow! Then here for a snack more faithful:

Answers:

X. = 6

X. = 13/31

X. = -0,75

X. 1 = 1; x. 2 = 8/3

And it is decided? Well, respect! I remove the hat.) So, the lesson passed not in vain, and the initial level of solution of the indicative equations can be considered successfully mastered. Ahead - the following levels and more complex equations! And new techniques and approaches. And non-standard examples. And new surprises.) All this is in the next lesson!

Something failed? So, most likely, problems in. Or in . Or in that and the other immediately. Here I'm powerless. I can once again offer only one thing - not to be lazy and stroll through the references.)

To be continued.)