Exponential equations in the exam profile level. Exponential Equations
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 understanding of the simplest equations - linear and square: $ 56x-11 \u003d $ 0; $ ((x) ^ (2)) + 5x + 4 \u003d 0 $; $ ((x) ^ (2)) - 12x + 32 \u003d 0 $, etc. It is absolutely necessary to be able to solve such constructions 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)) \u003d 4; \\ quad ((5) ^ (2x-3)) \u003d \\ frac (1) (25); \\ quad ((9) ^ (x)) \u003d - 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) \u003d ((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 exponential equations are much easier to give 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 brutal equations that solving them becomes problematic not only for students - even many teachers get stuck on such problems.
However, let's not talk about sad things. And 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)) \u003d 4 $. Well, to what degree should the number 2 be raised to get the number 4? Probably the second? After all, $ ((2) ^ (2)) \u003d 2 \\ cdot 2 \u003d 4 $ - and we got the correct numerical equality, i.e. really $ x \u003d 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)) \u003d \\ frac (1) (25) \\]
And here it is already a little more complicated. Many students know that $ ((5) ^ (2)) \u003d 25 $ is a multiplication table. Some also suspect that $ ((5) ^ (- 1)) \u003d \\ frac (1) (5) $ is essentially a definition of negative powers (similar to the formula $ ((a) ^ (- n)) \u003d \\ frac (1) (((a) ^ (n))) $).
Finally, only a select few guess that these facts can be combined and the following result can be obtained at the output:
\\ [\\ frac (1) (25) \u003d \\ frac (1) (((5) ^ (2))) \u003d ((5) ^ (- 2)) \\]
Thus, our original equation will be rewritten as follows:
\\ [((5) ^ (2x-3)) \u003d \\ frac (1) (25) \\ Rightarrow ((5) ^ (2x-3)) \u003d ((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 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 \u003d -2 \\\\ & 2x \u003d 3-2 \\\\ & 2x \u003d 1 \\\\ & x \u003d \\ 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 is too early for you to tackle the indicative equations.
\\ [((9) ^ (x)) \u003d - 3 \\]
Well, how to solve this? First thought: $ 9 \u003d 3 \\ cdot 3 \u003d ((3) ^ (2)) $, so the original equation can be rewritten like this:
\\ [((\\ left (((3) ^ (2)) \\ right)) ^ (x)) \u003d - 3 \\]
Then we remember that when raising a power to a power, the indicators are multiplied:
\\ [((\\ 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 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)) \u003d 3 & ((3) ^ (- 1)) \u003d \\ frac (1) (3) & ((3) ^ (\\ frac (1) ( 2))) \u003d \\ sqrt (3) \\\\ ((3) ^ (2)) \u003d 9 & ((3) ^ (- 2)) \u003d \\ frac (1) (9) & ((3) ^ (\\ 3) ^ (- \\ frac (1) (2))) \u003d \\ 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 gone! And it cannot be, because the exponential function $ y \u003d ((a) ^ (x)) $, firstly, always takes only positive values \u200b\u200b(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)) \u003d - 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)) \u003d 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)) \u003d 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 is any expression, we will get a new equation, which can already be solved. For instance:
\\ [\\ 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 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)) \u003d 3; \\ quad ((5) ^ (x)) \u003d 15; \\ quad ((4) ^ (2x)) \u003d 11 \\]
Well, to what degree should 2 be raised to get 3? First? But no: $ ((2) ^ (1)) \u003d 2 $ - not enough. The second? Also not: $ ((2) ^ (2)) \u003d 4 $ - a bit too much. Which one then?
Knowledgeable students have probably already guessed: in such cases, when it is impossible to solve “nicely”, “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)) \u003d 3 \\\\ & a \u003d ((b) ^ (((\\ log) _ (b)) a)) \\\\\\ end (align) \\]
If we assume that $ a \u003d 3 $ is our original number on the right, and $ b \u003d 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 \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) \\]
We got a slightly strange answer: $ x \u003d ((\\ log) _ (2)) 3 $. In some other task, many with such an answer 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)) \u003d 15 \\ Rightarrow ((5) ^ (x)) \u003d ((5) ^ (((\\ log) _ (5)) 15)) \\ Rightarrow x \u003d ((\\ log) _ (5)) 15; \\\\ & ((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 differently:
We introduced the factor 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)) \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 come across you very, 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? 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 and remember a couple of tricks 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:
- Write down the original equation. For example: $ ((4) ^ (x)) + ((4) ^ (x-1)) \u003d ((4) ^ (x + 1)) - 11 $;
- Make some kind of incomprehensible crap. Or even a few crap called "transform equation";
- At the output, get the simplest expressions like $ ((4) ^ (x)) \u003d 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, 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:
- The equation is composed of exponential functions with the same base. Example: $ ((4) ^ (x)) + ((4) ^ (x-1)) \u003d ((4) ^ (x + 1)) - 11 $;
- The formula contains exponential 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 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)) \u003d ((4) ^ (x + 1)) - 11 \\]
What do we see? The four is being raised to varying degrees. But all these powers are simple sums of the variable $ x $ with other numbers. Therefore, you need to remember 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))) (((a ) ^ (y))). \\\\\\ end (align) \\]
Simply put, addition of exponents can be converted to a product of powers, and subtraction can easily be converted to division. Let's try to apply these formulas to the powers 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. \\ 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) \u003d ((4) ^ (x)) \\ cdot 4 -eleven; \\\\ & ((4) ^ (x)) + ((4) ^ (x)) \\ cdot \\ frac (1) (4) - ((4) ^ (x)) \\ cdot 4 + 11 \u003d 0. \\\\\\ end (align) \\]
The first four terms contain the element $ ((4) ^ (x)) $ - take it outside the parenthesis:
\\ [\\ 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 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 ) \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 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 parenthesis) 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, let's analyze one more task:
\\ [((5) ^ (x + 2)) + ((0,2) ^ (- x-1)) + 4 \\ cdot ((5) ^ (x + 1)) \u003d 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)) \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)) ) \\]
{!LANG-46123ff60704d8c73690bbc4e3eb44b8!}
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)) \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, cheated a little. Because for a complete understanding, the formula for getting rid of negative indicators had to be written as follows:
\\ [((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 from working with only one fraction:
\\ [((\\ 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 raise the degree to a different degree (remember: in this case, the indicators add up). But I didn't have 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)) \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 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 reduced itself. It remains only to remember that $ 1 \u003d ((5) ^ (0)) $, whence we get:
\\ [\\ begin (align) & ((5) ^ (x + 2)) \u003d ((5) ^ (0)); \\\\ & x + 2 \u003d 0; \\\\ & x \u003d -2. \\\\\\ end (align) \\]
That's the whole solution! We got the final answer: $ x \u003d -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)) \u003d ((21) ^ (3x)); \\\\ & ((100) ^ (x-1)) \\ cdot ((2.7) ^ (1-x)) \u003d 0.09. \\\\\\ end (align) \\]
{!LANG-3219a6d83d2b7a9f94abbf6d558f5830!}
{!LANG-9803c2cf2500b8a8ec8e4af7795d1a5a!}
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)) \u003d ((21) ^ (3x)); \\\\ & 21 \u003d 7 \\ cdot 3 \\ Rightarrow ((21) ^ (3x)) \u003d ((\\ left (7 \\ cdot 3 \\ right)) ^ (3x)) \u003d ((7) ^ (3x)) \\ \\\\\\ end (align) \\]
But you can do the opposite - make the number 21 from the numbers 7 and 3. This is especially easy to do on the left, since the indicators of both degrees are the same:
\\ [\\ begin (align) & ((7) ^ (x + 6)) \\ cdot ((3) ^ (x + 6)) \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 moved 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)) \u003d 0.09 \\]
\\ [((100) ^ (x-1)) \\ cdot ((\\ left (\\ frac (27) (10) \\ right)) ^ (1-x)) \u003d \\ 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 with which to work.
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)) \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 moved the total exponent from the product outside the bracket according to the rule $ ((a) ^ (x)) \\ cdot ((b) ^ (x)) \u003d ((\\ 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. Than? Yes, it is obvious: they are powers 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 be rewritten 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)) \\]
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)) \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 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 it is necessary 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 independent 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. :)
On the youtube channel of our site, to keep abreast of all the new video lessons.
To begin with, let's recall the basic formulas of degrees and their properties.
Product of number a happens to itself n times, we can write this expression as a a ... a \u003d a n
1.a 0 \u003d 1 (a ≠ 0)
3.a n a m \u003d a n + m
4. (a n) m \u003d a nm
5.a n b n \u003d (ab) n
7.a n / a m \u003d a n - m
Power or exponential equations - these are equations in which the variables are in powers (or exponents), and the base is a number.
Examples of exponential equations:
In this example, the number 6 is the base, it always stands at the bottom, and the variable x degree or indicator.
Here are some more examples of exponential equations.
2 x * 5 \u003d 10
16 x - 4 x - 6 \u003d 0
Now let's see how the exponential equations are solved?
Let's take a simple equation:
2 x \u003d 2 3
Such an example can be solved even in the mind. It is seen that x \u003d 3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let's see how this solution needs to be formalized:
2 x \u003d 2 3
x \u003d 3
In order to solve such an equation, we removed identical grounds (that is, two's) and wrote down what was left, these are degrees. We got the desired answer.
Now let's summarize our decision.
Algorithm for solving the exponential equation:
1. Need to check the same whether the equation has bases on the right and left. If the grounds are not the same, we are looking for options to solve this example.
2. After the bases are the same, equate degree and solve the resulting new equation.
Now let's solve a few examples:
Let's start simple.
The bases on the left and right sides are equal to the number 2, so we can discard the base and equate their degrees.
x + 2 \u003d 4 This is the simplest equation.
x \u003d 4 - 2
x \u003d 2
Answer: x \u003d 2
In the following example, you can see that the bases are different - 3 and 9.
3 3x - 9x + 8 \u003d 0
First, we move the nine to the right side, we get:
Now you need to make the same bases. We know that 9 \u003d 3 2. Let's use the formula of degrees (a n) m \u003d a nm.
3 3x \u003d (3 2) x + 8
We get 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2x + 16
3 3x \u003d 3 2x + 16 now you can see that the bases on the left and right sides are the same and equal to three, so we can discard them and equate the degrees.
3x \u003d 2x + 16 got the simplest equation
3x - 2x \u003d 16
x \u003d 16
Answer: x \u003d 16.
See the following example:
2 2x + 4 - 10 4 x \u003d 2 4
First of all, we look at the bases, bases are different two and four. And we need to be - the same. We transform the four by the formula (a n) m \u003d a nm.
4 x \u003d (2 2) x \u003d 2 2x
And we also use one formula a n a m \u003d a n + m:
2 2x + 4 \u003d 2 2x 2 4
Add to the equation:
2 2x 2 4 - 10 2 2x \u003d 24
We have led the example to the same grounds. But we are hindered by other numbers 10 and 24. What to do with them? If you look closely, you can see that on the left side we repeat 2 2x, here is the answer - 2 2x we can take out of the brackets:
2 2x (2 4 - 10) \u003d 24
Let's calculate the expression in brackets:
2 4 — 10 = 16 — 10 = 6
Divide the whole equation by 6:
Let's imagine 4 \u003d 2 2:
2 2x \u003d 2 2 bases are the same, discard them and equate the powers.
2x \u003d 2 we get the simplest equation. We divide it by 2 we get
x \u003d 1
Answer: x \u003d 1.
Let's solve the equation:
9 x - 12 * 3 x + 27 \u003d 0
Let's transform:
9 x \u003d (3 2) x \u003d 3 2x
We get the equation:
3 2x - 12 3x +27 \u003d 0
Our bases are the same equal to 3. In this example, you can see that the first three has a degree twice (2x) than the second (just x). In this case, you can solve replacement method... We replace the number with the smallest degree:
Then 3 2x \u003d (3x) 2 \u003d t 2
Replace all powers with x in the equation with t:
t 2 - 12t + 27 \u003d 0
We get a quadratic equation. We solve through the discriminant, we get:
D \u003d 144-108 \u003d 36
t 1 \u003d 9
t 2 \u003d 3
Back to the variable x.
We take t 1:
t 1 \u003d 9 \u003d 3 x
That is,
3 x \u003d 9
3 x \u003d 3 2
x 1 \u003d 2
One root was found. We are looking for the second, from t 2:
t 2 \u003d 3 \u003d 3 x
3 x \u003d 3 1
x 2 \u003d 1
Answer: x 1 \u003d 2; x 2 \u003d 1.
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Lesson type
: a lesson of generalization and complex applications of knowledge, skills and abilities on the topic "Exponential equations and ways to solve them."Lesson objectives.
Equipment:
computer and multimedia projector.The lesson uses information Technology : methodological support for the lesson - presentation in Microsoft Power Point program.
During the classes
Every skill is given by labor
I. Lesson goal setting(Slide number 2 )
In this lesson, we will summarize and generalize the topic “Exponential equations, their solutions”. Let's get acquainted with typical USE assignments from different years on this topic.
Tasks for solving exponential equations can be found in any part of the exam tasks. In the part “ IN " usually they offer to solve the simplest exponential equations. In the part “ FROM " you can find more complex exponential equations, the solution of which is usually one of the stages of the task.
For instance ( Slide number 3 ).
- Unified State Exam - 2007
Q 4 - Find the largest expression value x ywhere ( x; at) - system solution:
B 1 - Solve equations:
and) x 6 3x – 36 6 3x = 0;
b) 4 x +1 + 8 4 x= 3.
Q 4 - Find the meaning of the expression x + ywhere ( x; at) - system solution:
- Unified State Exam - 2010
II. Updating basic knowledge. Reiteration
(Slides number 4 - 6 presentations for the lesson)The screen shows basic summary of theoretical material on this topic.
The following issues are discussed:
- What equations are called indicative?
- Name the main ways to solve them. Give examples of their types ( Slide number 4 )
- Which theorem is used to solve the simplest exponential equations of the form: and f (x) \u003d a g (x)?
- What other methods for solving exponential equations are there? ( Slide number 5 )
(Solve the proposed equations for each method on your own and do a self-test using a slide)
- Factoring method (based on properties of degrees with the same bases, admission: the degree with the smallest exponent is taken out of the bracket).
- Reception of division (multiplication) by an exponential expression other than zero, when solving homogeneous exponential equations .
- Advice:
-
Solving equations with the last two methods with subsequent comments
(Slide number 6 ).
. 4 x+ 1 – 2 4 x– 2 = 124, 4 x– 2 (4 3 - 2) = 124, 4 x– 2 62 = 124,4 x– 2 = 2, 4 x– 2 = 4 0,5 , x– 2 = 0,5, x \u003d 2,5 .
2 2 2x - 3 2 x 5 x - 5 5 2x \u003d 0¦: 5 2 x0,2 (2/5) 2x - 3 (2/5) x - 5 = 0,
t \u003d (2/5) x, t > 0, 2t 2 - 3 t - 5 = 0, t= -1(?...), t \u003d 5/2; 5/2 \u003d (2/5) x, x= ?...
III. Solving the tasks of the exam 2010
Students independently solve the tasks proposed at the beginning of the lesson on slide number 3, using the directions for the solution, check their solution course and answers to them using the presentation ( Slide number 7 ). In the course of work, options and methods of solution are discussed, attention is drawn to possible errors in the solution.
: a) 7 x- 2 \u003d 49, b) (1/6) 12 - 7 x = 36. Answer: and) x\u003d 4, b) x = 2. : 4 x2 + 3x – 2 - 0,5 2x2 + 2x - 1 \u003d 0. (You can replace 0.5 \u003d 4 - 0.5)Decision. ,
x 2 + 3x – 2 = -x 2 - 4x + 0,5 …
Answer: x= -5/2, x = 1/2.
: 5 5 tg y + 4 \u003d 5 -tg y , with cos y< 0.Indication to the solution
... 5 5 tg y + 4 \u003d 5 -tg y ¦ 5 tg y 0,5 5 2g y + 4 5 tg y - 1 \u003d 0. Let x\u003d 5 tg y , …
5 tg y = -1 (?...), 5 tg y \u003d1/5.
Since tg y\u003d -1 and cos y< 0, then at II coordinate quarter
Answer: at= 3/4 + 2k, k N.
IV. Collaborate at the blackboard
The task of a high level of training is considered - Slide number 8 ... With the help of this slide, a dialogue between teacher and students takes place, contributing to the development of the solution.
- At what parameter and equation 2 2 x – 3 2 x + and 2 – 4and \u003d 0 has two roots?Let be t= 2 x where t > 0 ... We get t 2 – 3t + (and 2 – 4and) = 0 .
1). Since the equation has two roots, D\u003e 0;
2). Because t 1,2\u003e 0, then t 1 t 2\u003e 0, that is and 2 – 4and> 0 (?...).
Answer: and(- 0.5; 0) or (4; 4.5).
V. Verification work
(Slide number 9 )Students perform verification work on pieces of paper, exercising self-control and self-assessment of the work performed with the help of a presentation, affirming the topic. They independently determine for themselves a program for regulating and correcting knowledge based on mistakes made in workbooks. Sheets with completed independent work are handed over to the teacher for verification.
Underlined numbers - basic level, with an asterisk - increased difficulty.
Solution and answers.
3. 2 x– 1 (5 2 4 - 4) = 19, 2 x– 1 76 = 19, 2 x– 1 = 1/4, 2 x– 1 = 2 – 2 , x– 1 = -2,
x \u003d -1.
4 * .3 9 x \u003d 2 3 x 5 x+ 5 25 x | : 25 x ,
3 (9/25) x \u003d 2 (3/5) x+ 5,
3 (9/27) x = 2 (3/5) x + 5 = 0,
3 (3/5) 2x – 2 (3/5) x - 5 = 0,…, (3/5) x = -1 (does not fit),
(3/5) x = 5, x \u003d -1.
Vi. Homework
(Slide number 10 )- Repeat § 11, 12.
- From the materials of the Unified State Exam 2008 - 2010, select tasks on the topic and solve them.
- Home test work :
Do not be intimidated by my words, you have already encountered this method in the 7th grade, when you studied polynomials.
For example, if you needed:
Let's group it up: the first and third terms, as well as the second and fourth.
It is clear that the first and third are the difference of squares:
and the second and fourth have a common factor of three:
Then the original expression is equivalent to this:
Where to take out the common factor is no longer difficult:
Hence,
This is approximately how we will act when solving exponential equations: look for "commonality" among the terms and put it outside the brackets, well then - come what may, I believe that we will be lucky \u003d))
Example No. 14
On the right is far from a degree of seven (I checked it!) And on the left - not much better ...
You can, of course, "chop off" the multiplier a from the second from the first term, and then deal with the result, but let's do it more sensibly with you.
I don’t want to deal with fractions, which inevitably come from “selection,” so wouldn't it be better for me to endure?
Then I won't have fractions: as they say, both the wolves are fed and the sheep are safe:
Count the expression in parentheses.
In a magical, magical way, it turns out that (surprising, although what else can we expect?).
Then we will cancel both sides of the equation by this factor. We get:, from where.
Here is a more complicated example (quite a bit, really):
What a trouble! We don't have one common ground here!
It is not entirely clear what to do now.
Let's do what we can: first, let's move the "fours" to one side, and the "fives" to the other:
Now let's move the "common" to the left and right:
So what now?
What is the benefit of such a stupid group? At first glance, it is not visible at all, but let's look deeper:
Well, now let's make it so that on the left we have only the expression with, and on the right - everything else.
How do we do this?
Here's how: Divide both sides of the equation first by (this way we get rid of the power on the right), and then divide both sides by (this way we get rid of the numerical factor on the left).
We finally get:
Incredible!
On the left we have an expression, and on the right we have a simple one.
Then we immediately conclude that
Example No. 15
I will give his brief solution (without bothering too much with explanations), try to figure out all the "subtleties" of the solution yourself.
Now the final consolidation of the passed material.
Solving the following 7 problems independently (with answers)
- Let's take the common factor out of the brackets:
- We represent the first expression in the form:, divide both parts into and get that
- , then the original equation is transformed to the form: Well, now a hint - look for where you and I have already solved this equation!
- Imagine how, how, and, well, then divide both parts by, so you get the simplest exponential equation.
- Take out the brackets.
- Take out the brackets.
EXPLORATIVE EQUATIONS. AVERAGE LEVEL
I assume that after reading the first article that told what are exponential equations and how to solve them, you have mastered the necessary minimum of knowledge required to solve the simplest examples.
Now I will analyze another method for solving exponential equations, this ...
Method of introducing a new variable (or replacement)
He solves most of the "difficult" problems on the topic of exponential equations (and not only equations).
This method is one of most often used in practice. First, I recommend that you familiarize yourself with the topic.
As you already understood from the name, the essence of this method is to introduce such a change of variable that your exponential equation miraculously transforms into one that you can easily solve.
All that is left for you after solving this very “simplified equation” is to make a “reverse replacement”: that is, to return from the replaced to the replaced one.
Let's illustrate what we just said with a very simple example:
Example 16. Simple replacement method
This equation is solved using "Simple replacement", as mathematicians scornfully call it.
Indeed, the replacement is the most obvious here. One has only to see that
Then the original equation will turn into this:
If you additionally imagine how, then it is quite clear what needs to be replaced ...
Of course, .
What then will the original equation turn into? And here's what:
You can easily find its roots yourself:.
What should we do now?
It's time to go back to the original variable.
What did I forget to indicate?
Namely: when replacing some degree with a new variable (that is, when changing the view), I will be interested in only positive roots!
You yourself can easily answer why.
Thus, you and I are not interested, but the second root is quite suitable for us:
Then where.
Answer:
As you can see, in the previous example, the replacement was asking for our hands. Unfortunately, this is not always the case.
However, let's not go straight to the sad, but practice with one more example with a fairly simple replacement
Example 17. Simple replacement method
It is clear that most likely it will have to be replaced (this is the smallest of the degrees included in our equation).
However, before introducing the replacement, our equation must be "prepared" for it, namely:,.
Then you can replace, as a result I get the following expression:
Oh horror: a cubic equation with completely creepy formulas for its solution (well, speaking in general terms).
But let's not despair right away, but think about what to do.
I will propose to cheat: we know that in order to get a “nice” answer, we need to get it in the form of some power of a triple (why would that be, eh?).
Let's try to guess at least one root of our equation (I'll start guessing with powers of three).
First assumption. It is not a root. Alas and ah ...
.
The left side is equal.
Right part: !
There is! You guessed the first root. Now things will get easier!
Do you know about the “corner” division scheme? Of course you know you use it when you divide one number by another.
But few people know that the same can be done with polynomials.
There is one great theorem:
Applicable to my situation, this tells me what is divisible by.
How is division carried out? That's how:
I look at which monomial I have to multiply to get
It is clear that on, then:
Subtract the resulting expression from, get:
Now what do I need to multiply by to get?
It is clear that on, then I will get:
and again subtract the resulting expression from the remaining one:
Well, the last step, I will multiply by, and subtract from the remaining expression:
Hurray, the division is over! What have we saved up in private?
By itself: .
Then we got the following decomposition of the original polynomial:
Let's solve the second equation:
It has roots:
Then the original equation:
has three roots:
We will, of course, discard the last root, since it is less than zero.
And the first two after the reverse replacement will give us two roots:
Answer: ..
I didn’t mean to scare you with this example!
On the contrary, my goal was to show that even though we had a fairly simple replacement, it nevertheless led to a rather complex equation, the solution of which required some special skills from us.
Well, no one is immune from this. But the replacement in this case was pretty obvious.
Example # 18 (with a less obvious replacement)
It is not at all clear what we should do: the problem is that in our equation there are two different bases and one base cannot be obtained from the other by raising to any (reasonable, naturally) degree.
However, what do we see?
Both bases differ only in sign, and their product is the difference of squares equal to one:
Definition:
Thus, the numbers that are the bases in our example are conjugate.
In this case, a smart move would be multiply both sides of the equation by the conjugate number.
For example, on, then the left side of the equation becomes equal, and the right.
If we make a substitution, then our original equation will become like this:
its roots, then, and remembering that, we get that.
Answer:,.
As a rule, the replacement method is sufficient to solve most of the "school" exponential equations.
The following tasks of an increased level of complexity are taken from the USE versions.
Three tasks of increased complexity from the options of the exam
You are already competent enough to independently solve these examples. I will only give the required replacement.
- Solve the equation:
- Find the roots of the equation:
- Solve the equation:. Find all the roots of this equation that belong to the segment:
And now, brief explanations and answers:
Example No. 19
Here it is enough for us to note that and.
Then the original equation will be equivalent to this:
This equation is solved by replacing
Do the further calculations yourself.
In the end, your task will be reduced to solving the simplest trigonometric (depending on sine or cosine). We will analyze the solution of such examples in other sections.
Example No. 20
You can even do without replacement here ...
It is enough to move the subtracted to the right and represent both bases through powers of two:, and then go directly to the quadratic equation.
Example No. 21
It is also solved in a fairly standard way: imagine how.
Then replacing we get a quadratic equation: then,
You already know what a logarithm is? No? Then read the topic urgently!
The first root, obviously, does not belong to the segment, and the second is incomprehensible!
But we'll find out very soon!
Since, then (this is a property of the logarithm!)
Subtract from both parts, then we get:
The left side can be represented as:
we multiply both parts by:
can be multiplied by, then
Then let's compare:
since, then:
Then the second root belongs to the required interval
Answer:
As you see, the selection of roots of exponential equations requires a sufficiently deep knowledge of the properties of logarithmsso I advise you to be as careful as possible when solving the exponential equations.
As you can imagine, in mathematics, everything is interconnected!
As my math teacher used to say: "math, like history, you can't read overnight."
As a rule, all the difficulty in solving problems of an increased level of complexity is precisely the selection of the roots of the equation.
Another example for training ...
Example 22
It is clear that the equation itself is quite simple to solve.
By making the substitution, we will reduce our original equation to the following:
First let's consider first root.
Let's compare and: since, then. (property of the logarithmic function, at).
Then it is clear that the first root does not belong to our interval either.
Now the second root:. It is clear that (since the function at is increasing).
It remains to compare and.
since, then, at the same time.
This way, I can drive a peg between and.
This peg is a number.
The first expression is smaller and the second is larger.
Then the second expression is greater than the first and the root belongs to the interval.
Answer:.
Finally, let's look at another example of an equation where the substitution is quite non-standard.
Example No. 23 (Equation with non-standard substitution!)
Let's start right away with what you can do, and what you can do, but it's better not to do it.
You can - represent everything through powers of three, two and six.
Where it leads?
And it will not lead to anything: a hodgepodge of degrees, and some of them will be quite difficult to get rid of.
What then is needed?
Let's note that a
And what will it give us?
And the fact that we can reduce the solution of this example to the solution of a fairly simple exponential equation!
First, let's rewrite our equation as:
Now we divide both sides of the resulting equation by:
Eureka! Now we can replace, we get:
Well, now it's your turn to solve demonstration problems, and I will give only brief comments to them, so that you do not go astray! Good luck!
Example No. 24
The most difficult!
It’s not easy to find a replacement here! Nevertheless, we can completely solve this example using selection of a full square.
To solve it, it is enough to note that:
Then here's a replacement for you:
(Please note that here, during our replacement, we cannot drop the negative root !!! And why do you think?)
Now, to solve the example, you have to solve two equations:
Both of them are solved by the "standard replacement" (but the second in one example!)
Example No. 25
2. Note that and make a replacement.
Example No. 26
3. Decompose the number into coprime factors and simplify the resulting expression.
Example No. 27
4. Divide the numerator and denominator of the fraction by (or, if you prefer) and replace or.
Example No. 28
5. Note that the numbers and are conjugate.
SOLUTION OF EXPRESS EQUATIONS BY THE LOGARITHM METHOD. ADVANCED LEVEL
In addition, let's consider another way - solution of exponential equations by the logarithm method.
I can't say that the solution of exponential equations by this method is very popular, but in some cases only it can lead us to the correct solution of our equation.
It is especially often used to solve the so-called " mixed equations»: That is, those where functions of different types meet.
Example No. 29
in the general case, it can only be solved by taking the logarithm of both sides (for example, by the base), in which the original equation turns into the following:
Let's consider the following example:
It is clear that according to the ODZ of the logarithmic function, we are only interested in.
However, this follows not only from the ODZ of the logarithm, but for another reason.
I think that it will not be difficult for you to guess which one.
Let's log both sides of our equation to the base:
As you can see, taking the logarithm of our original equation quickly enough led us to the correct (and beautiful!) Answer.
Let's practice with one more example.
Example No. 30
Here, too, there is nothing wrong: we logarithm both sides of the equation by the base, then we get:
Let's make a replacement:
However, we are missing something! Have you noticed where I went wrong? After all, then:
which does not satisfy the requirement (think where it came from!)
Answer:
Try to write down the solution of the exponential equations below yourself:
Now check your decision against this:
Example No. 31
Logarithm both sides to the base, taking into account that:
(the second root does not suit us due to the replacement)
Example No. 32
Logarithm base:
Let's transform the resulting expression to the following form:
EXPLORATIVE EQUATIONS. BRIEF DESCRIPTION AND BASIC FORMULAS
Exponential equation
Equation of the form:
called the simplest exponential equation.
Power properties
Solution approaches
- Coercing to the same base
- Conversion to the same exponent
- Variable replacement
- Simplification of expression and application of one of the above.
