Unified State Exam prototypes exponential equations test options. Exponential equations. Logarithm method. Replacing a variable in solving exponential equations. Examples
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First, let's remember the basic formulas of powers and their properties.
Product of a number a occurs on itself n times, we can write this expression as a a … a=a n
1. a 0 = 1 (a ≠ 0)
3. a n a m = a n + m
4. (a n) m = a nm
5. a n b n = (ab) n
7. a n / a m = 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 is always at the bottom, and the variable x degree or indicator.
Let us give more examples of exponential equations.
2 x *5=10
16 x - 4 x - 6=0
Now let's look at how exponential equations are solved?
Let's take a simple equation:
2 x = 2 3
This example can be solved even in your head. It can be seen that x=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 to formalize this decision:
2 x = 2 3
x = 3
In order to solve such an equation, we removed identical grounds(that is, twos) and wrote down what was left, these are degrees. We got the answer we were looking for.
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 reasons are not the same, we are looking for options to solve this example.
2. After the bases become the same, equate degrees and solve the resulting new equation.
Now let's look at a few examples:
Let's start with something simple.
The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their degrees.
x+2=4 The simplest equation is obtained.
x=4 – 2
x=2
Answer: x=2
In the following example you can see that the bases are different: 3 and 9.
3 3x - 9 x+8 = 0
First, move the nine to the right side, we get:
Now you need to make the same bases. We know that 9=3 2. Let's use the power formula (a n) m = a nm.
3 3x = (3 2) x+8
We get 9 x+8 =(3 2) x+8 =3 2x+16
3 3x = 3 2x+16 Now it is clear that on the left and right sides the bases are the same and equal to three, which means we can discard them and equate the degrees.
3x=2x+16 we get the simplest equation
3x - 2x=16
x=16
Answer: x=16.
Let's look at the following example:
2 2x+4 - 10 4 x = 2 4
First of all, we look at the bases, bases two and four. And we need them to be the same. We transform the four using the formula (a n) m = a nm.
4 x = (2 2) x = 2 2x
And we also use one formula a n a m = a n + m:
2 2x+4 = 2 2x 2 4
Add to the equation:
2 2x 2 4 - 10 2 2x = 24
We gave an example for the same reasons. But other numbers 10 and 24 bother us. What to do with them? If you look closely you can see that on the left side we have 2 2x repeated, here is the answer - we can put 2 2x out of brackets:
2 2x (2 4 - 10) = 24
Let's calculate the expression in brackets:
2 4 — 10 = 16 — 10 = 6
We divide the entire equation by 6:
Let's imagine 4=2 2:
2 2x = 2 2 bases are the same, we discard them and equate the degrees.
2x = 2 is the simplest equation. Divide it by 2 and we get
x = 1
Answer: x = 1.
Let's solve the equation:
9 x – 12*3 x +27= 0
Let's convert:
9 x = (3 2) x = 3 2x
We get the equation:
3 2x - 12 3 x +27 = 0
Our bases are the same, equal to three. 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 = (3 x) 2 = t 2
We replace all x powers in the equation with t:
t 2 - 12t+27 = 0
We get a quadratic equation. Solving through the discriminant, we get:
D=144-108=36
t 1 = 9
t2 = 3
Returning to the variable x.
Take t 1:
t 1 = 9 = 3 x
That is,
3 x = 9
3 x = 3 2
x 1 = 2
One root was found. We are looking for the second one from t 2:
t 2 = 3 = 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 = 2; x 2 = 1.
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Don't be scared by my words, you already came across this method in 7th grade when you studied polynomials.
For example, if you needed:
Let's group: the first and third terms, as well as the second and fourth.
It is clear that the first and third are the difference of squares:
and the second and fourth have common multiplier three:
Then the original expression is equivalent to this:
Where to derive the common factor is no longer difficult:
Hence,
This is roughly what we will do when solving exponential equations: look for “commonality” among the terms and take it out of brackets, and then - come what may, I believe that we will be lucky =))
Example No. 14
The right is far from a power of seven (I checked!) And the left is not much better...
You can, of course, “chop off” the factor a from the second term from the first term, and then deal with what you got, but let’s be more prudent with you.
I don't want to deal with the fractions that inevitably form when "selecting" , so shouldn't I rather take it out?
Then I won’t have any fractions: as they say, the wolves are fed and the sheep are safe:
Calculate the expression in brackets.
Magically, magically, it turns out that (surprisingly, although what else should we expect?).
Then we reduce both sides of the equation by this factor. We get: , from.
Here's a more complicated example (quite a bit, really):
What a problem! We don't have one common ground here!
It's not entirely clear what to do now.
Let’s do what we can: first, move the “fours” to one side, and the “fives” to the other:
Now let's take out the "general" on the left and right:
So what now?
What is the benefit of such a stupid group? At first glance it is not visible at all, but let's look deeper:
Well, now we’ll make sure that on the left we only have the expression c, and on the right - everything else.
How do we do this?
Here's how: Divide both sides of the equation first by (so we get rid of the exponent on the right), and then divide both sides by (so we get rid of the numeric factor on the left).
Finally we get:
Incredible!
On the left we have an expression, and on the right we have a simple expression.
Then we immediately conclude that
Example No. 15
I'll bring him short solution(without really bothering yourself with explanations), try to understand all the “subtleties” of the solution yourself.
Now for the final consolidation of the material covered.
Independently solving the following 7 problems (with answers)
- Let's take the common factor out of brackets: Where:
- Let's present the first expression in the form: , divide both sides by 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, ah, well, then divide both sides by, so you get the simplest exponential equation.
- Bring it out of the brackets.
- Bring it out of the brackets.
EXPONENTARY EQUATIONS. AVERAGE LEVEL
I assume that after reading the first article, which talked about what are exponential equations and how to solve them, you have mastered the necessary minimum knowledge necessary to solve the simplest examples.
Now I will look at another method for solving exponential equations, this is...
Method for introducing a new variable (or replacement)
He solves most “difficult” problems on the topic of exponential equations (and not only equations).
This method is one of most frequently used in practice. First, I recommend that you familiarize yourself with the topic.
As you already understood from the name, the essence of this method is to introduce such a change of variable that your exponential equation will miraculously transform into one that you can easily solve.
All that remains for you after solving this very “simplified equation” is to make a “reverse replacement”: that is, return from the replaced to the replaced.
Let's illustrate what we just said with a very simple example:
Example 16. Simple replacement method
This equation can be solved using "simple replacement", as mathematicians disparagingly call it.
In fact, the replacement here is the most obvious. One has only to see that
Then the original equation will turn into this:
If you additionally imagine how, then it is absolutely clear that it is necessary to replace...
Of course, .
What then becomes the original equation? Here's what:
You can easily find its roots on your own: .
What should we do now?
It's time to return to the original variable.
What did I forget to mention?
Namely: when replacing a certain degree with a new variable (that is, when replacing a type), I will be interested in only positive roots!
You yourself can easily answer why.
Thus, you and I are not interested, but the second root is quite suitable for us:
Then where from.
Answer:
As you can see, in the previous example, a replacement was just asking for our hands. Unfortunately, this is not always the case.
However, let’s not go straight to the sad stuff, but let’s practice with one more example with a fairly simple replacement
Example 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 a replacement, our equation needs to be “prepared” for it, namely: , .
Then you can replace, as a result I get the following expression:
Oh horror: a cubic equation with absolutely terrible formulas for solving it (well, speaking in general terms).
But let’s not despair right away, but let’s think about what we should do.
I'll suggest cheating: we know that to get a “beautiful” answer, we need to get it in the form of some power of three (why would that be, eh?).
Let's try to guess at least one root of our equation (I'll start guessing with powers of three).
First guess. Not a root. Alas and ah...
.
The left side is equal.
Right part: !
Eat! Guessed the first root. Now things will get easier!
Do you know about the “corner” division scheme? Of course you do, you use it when you divide one number by another.
But few people know that the same can be done with polynomials.
There is one wonderful theorem:
Applying to my situation, this tells me that it is divisible without remainder by.
How is division carried out? That's how:
I look to see which monomial I should multiply by to get
It is clear that on, then:
I subtract the resulting expression from, I get:
Now, what do I need to multiply by to get?
It is clear that on, then I will get:
and again subtract the resulting expression from the remaining one:
Well, the last step is to multiply by and subtract from the remaining expression:
Hurray, division is over! What have we accumulated in private?
By itself: .
Then we got the following expansion of the original polynomial:
Let's solve the second equation:
It has roots:
Then the original equation:
has three roots:
We will, of course, discard the last root, since it is less than zero.
And the first two after reverse replacement will give us two roots:
Answer: ..
I didn’t mean to scare you with this example!
Rather, on the contrary, my goal was to show that although 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 in this case was pretty obvious.
Example No. 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 it to any (reasonable, naturally) power.
However, what do we see?
Both bases differ only in sign, and their product is the difference of squares equal to one:
Definition:
Thus, the numbers that are the bases in our example are conjugate.
In this case, the smart step would be Multiply both sides of the equation by the conjugate number.
For example, on, then the left side of the equation will become equal to, and the right.
If we make a substitution, then our original equation will become like this:
its roots, then, and remembering that, we get that.
Answer: , .
As a rule, the replacement method is sufficient to solve most “school” exponential equations.
The following tasks of an increased level of complexity are taken from the Unified State Exam variants.
Three tasks of increased complexity from the Unified State Exam variants
You are already literate enough to solve these examples on your own. 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 some brief explanations and answers:
Example No. 19
Here it is enough for us to note that...
Then the original equation will be equivalent to this:
This equation can be solved by replacing
Do the further calculations yourself.
In the end, your task will be reduced to solving simple trigonometric problems (depending on sine or cosine). We will look at solutions to similar examples in other sections.
Example No. 20
Here you can even do without replacement...
It is enough to move the subtrahend to the right and represent both bases through powers of two: , and then immediately move on to the quadratic equation.
Example No. 21
This is also solved in a fairly standard way: let’s imagine how.
Then, replacing, we get a quadratic equation: then,
You already know what a logarithm is, right? No? Then read the topic urgently!
The first root obviously does not belong to the segment, but the second one is unclear!
But we will find out very soon!
Since, then (this is a property of the logarithm!)
Subtract from both sides, then we get:
The left side can be represented as:
multiply both sides by:
can be multiplied by, then
Then compare:
since then:
Then the second root belongs to the required interval
Answer:
As you see, selection of roots of exponential equations requires sufficient deep knowledge properties of logarithms, so I advise you to be as careful as possible when solving exponential equations.
As you understand, in mathematics everything is interconnected!
As my math teacher said: “mathematics, like history, cannot be read overnight.”
As a rule, all The difficulty in solving problems of an increased level of complexity is precisely the selection of the roots of the equation.
Another example for practice...
Example 22
It is clear that the equation itself is solved quite simply.
By making a substitution, we reduce our original equation to the following:
First let's look at first root.
Let's compare and: since, then. (property logarithmic function, at).
Then it is clear that the first root does not belong to our interval.
Now the second root: . It is clear that (since the function at is increasing).
It remains to compare and...
since, then, at the same time.
This way I can “drive a peg” between the and.
This peg is a number.
The first expression is less and the second is greater.
Then the second expression is greater than the first and the root belongs to the interval.
Answer: .
Finally, let's look at another example of an equation where the substitution is quite unusual.
Example No. 23 (Equation with non-standard replacement!)
Let's start right away with what can be done, and what - in principle, can be done, but it is better not to do it.
You can imagine everything through the powers of three, two and six.
Where it leads?
It won’t lead to anything: a jumble of degrees, some of which will be quite difficult to get rid of.
What then is needed?
Let's notice that a
And what will this give us?
And the fact that we can reduce the solution of this example to the solution of a fairly simple exponential equation!
First, let's rewrite our equation as:
Now let's divide both sides of the resulting equation by:
Eureka! Now we can replace, we get:
Well, now it’s your turn to solve demonstration problems, and I will give only brief comments to them so that you don’t go astray! Good luck!
Example No. 24
The most difficult!
It’s so hard to see a replacement here! But nevertheless, this example can be completely solved using highlighting a complete square.
To solve it, it is enough to note that:
Then here's your replacement:
(Please note that here during our replacement we cannot discard the negative root!!! Why do you think?)
Now to solve the example you only have to solve two equations:
Both of them can be solved by a “standard replacement” (but the second one in one example!)
Example No. 25
2. Notice 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 make the substitution or.
Example No. 28
5. Notice that the numbers and are conjugate.
SOLVING EXPONENTARY EQUATIONS USING THE LOGARIFHM METHOD. ADVANCED LEVEL
In addition, let's look at another way - solving exponential equations using the logarithm method.
I can’t say that solving exponential equations using this method is very popular, but in some cases only it can lead us to the right decision our equation.
It is especially often used to solve the so-called “ mixed equations": that is, those where functions of different types occur.
Example No. 29
in the general case, it can only be solved by taking logarithms of both sides (for example, to the base), in which the original equation will turn into the following:
Let's look at the following example:
It is clear that according to the ODZ of the logarithmic function, we are only interested.
However, this follows not only from the ODZ of the logarithm, but for one more reason.
I think it won’t be difficult for you to guess which one it is.
Let's take the logarithm of both sides of our equation to the base:
As you can see, taking the logarithm of our original equation quickly led us to the correct (and beautiful!) answer.
Let's practice with one more example.
Example No. 30
There’s nothing wrong here either: let’s take the logarithm of both sides of the equation to the base, then we get:
Let's make a replacement:
However, we missed something! Did you notice where I made a mistake? After all, then:
which does not satisfy the requirement (think where it came from!)
Answer:
Try to write down the solution to the exponential equations below:
Now compare your decision with this:
Example No. 31
Let's logarithm both sides to the base, taking into account that:
(the second root is not suitable for us due to replacement)
Example No. 32
Let's take logarithms to the base:
Let us transform the resulting expression to the following form:
EXPONENTARY EQUATIONS. BRIEF DESCRIPTION AND BASIC FORMULAS
Exponential equation
Equation of the form:
called the simplest exponential equation.
Properties of degrees
Approaches to solution
- Reduction to the same basis
- Leading to the same indicator degrees
- Variable replacement
- Simplifying the expression and applying one of the above.
At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “Exponential Equations.” The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of problem, graduates can count on high scores when passing the Unified State Exam in mathematics.
Get ready for exam testing with Shkolkovo!
When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. A school textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.
The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.
Shkolkovo teachers collected, systematized and presented everything necessary for successful completion Unified State Exam material in the simplest and most accessible form.
Basic definitions and formulas are presented in the “Theoretical background” section.
To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples of exponential equations with solutions presented on this page to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or go straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.
Those examples with indicators that caused you difficulties can be added to “Favorites”. This way you can quickly find them and discuss the solution with your teacher.
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This lesson is intended for those who are just beginning to learn exponential equations. As always, let's start with the 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 quadratic: $56x-11=0$; $((x)^(2))+5x+4=0$; $((x)^(2))-12x+32=0$, etc. Being able to solve such constructions is absolutely necessary in order not to “get stuck” in the topic that will now be discussed.
So, exponential equations. Let me give you a couple of examples:
\[((2)^(x))=4;\quad ((5)^(2x-3))=\frac(1)(25);\quad ((9)^(x))=- 3\]
Some of them may seem more complex to you, while others, on the contrary, are too simple. But they all have one important feature in common: their notation contains the exponential function $f\left(x \right)=((a)^(x))$. Thus, let's introduce the definition:
An exponential equation is any equation containing an exponential function, i.e. expression of the form $((a)^(x))$. In addition to the indicated function, such equations can contain any other algebraic constructions - polynomials, roots, trigonometry, logarithms, etc.
OK then. We've sorted 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 teaching many students, I can say that most of them find exponential equations much easier than the same logarithms, and even more so trigonometry.
But there is bad news: sometimes the writers of problems for all kinds of textbooks and exams are struck by “inspiration”, and their drug-inflamed brain begins to produce 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 let's return to those three equations that were given at the very beginning of the story. Let's try to solve each of them.
First equation: $((2)^(x))=4$. Well, to what power must you raise the number 2 to get the number 4? Probably the second? After all, $((2)^(2))=2\cdot 2=4$ - and we got the correct numerical equality, i.e. indeed $x=2$. Well, thanks, Cap, but this equation was so simple that even my cat could solve it. :)
Let's look at the following equation:
\[((5)^(2x-3))=\frac(1)(25)\]
But here it’s a little more complicated. Many students know that $((5)^(2))=25$ is the multiplication table. Some also suspect that $((5)^(-1))=\frac(1)(5)$ is essentially the definition negative powers(by analogy with the formula $((a)^(-n))=\frac(1)(((a)^(n)))$).
Finally, only a select few realize that these facts can be combined and yield the following result:
\[\frac(1)(25)=\frac(1)(((5)^(2)))=((5)^(-2))\]
Thus, our original equation will be rewritten as follows:
\[((5)^(2x-3))=\frac(1)(25)\Rightarrow ((5)^(2x-3))=((5)^(-2))\]
But this is already completely solvable! On the left in the equation there is an exponential function, on the right in the equation there is an exponential function, there is nothing else anywhere except them. Therefore, we can “discard” the bases and stupidly equate the indicators:
We have obtained the simplest linear equation that any student can solve in just a couple of lines. Okay, in four lines:
\[\begin(align)& 2x-3=-2 \\& 2x=3-2 \\& 2x=1 \\& x=\frac(1)(2) \\\end(align)\]
If you don’t 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 take on exponential equations.
\[((9)^(x))=-3\]
So how can we solve this? First thought: $9=3\cdot 3=((3)^(2))$, so the original equation can be rewritten as follows:
\[((\left(((3)^(2)) \right))^(x))=-3\]
Then we remember that when raising a power to a power, the exponents are multiplied:
\[((\left(((3)^(2)) \right))^(x))=((3)^(2x))\Rightarrow ((3)^(2x))=-(( 3)^(1))\]
\[\begin(align)& 2x=-1 \\& x=-\frac(1)(2) \\\end(align)\]
And for such a decision we will receive a honestly deserved two. For, with the equanimity of a Pokemon, we sent the minus sign in front of the three to the power of this very three. But you can’t do that. And that's why. Take a look at different degrees triplets:
\[\begin(matrix) ((3)^(1))=3& ((3)^(-1))=\frac(1)(3)& ((3)^(\frac(1)( 2)))=\sqrt(3) \\ ((3)^(2))=9& ((3)^(-2))=\frac(1)(9)& ((3)^(\ frac(1)(3)))=\sqrt(3) \\ ((3)^(3))=27& ((3)^(-3))=\frac(1)(27)& (( 3)^(-\frac(1)(2)))=\frac(1)(\sqrt(3)) \\\end(matrix)\]
When compiling this tablet, I did not pervert anything: I looked at positive powers, and negative ones, and even fractional ones... well, where is at least one negative number here? He's gone! And it cannot be, because the exponential function $y=((a)^(x))$, firstly, always takes only positive values (no matter how much one is multiplied or divided by two, it will still be a positive number), and secondly, the base of such a function - the number $a$ - is by definition a positive number!
Well, how then to solve the equation $((9)^(x))=-3$? But no way: there are no roots. And in this sense, exponential equations are very similar to quadratic equations - there may also be no roots. But if in quadratic equations the number of roots is determined by the discriminant (positive discriminant - 2 roots, negative - no roots), then in exponentials everything depends on what is to the right of the equal sign.
Thus, we formulate the key conclusion: the simplest exponential equation of the form $((a)^(x))=b$ has a root if and only if $b \gt 0$. Knowing this simple fact, you can easily determine whether the equation proposed to you has roots or not. Those. Is it worth solving it at all or immediately writing down that there are no roots.
This knowledge will help us many times when we have to decide more complex tasks. For now, enough of the lyrics - it’s time to study the basic algorithm for solving exponential equations.
How to Solve Exponential Equations
So, let's formulate the problem. It is necessary to solve the exponential equation:
\[((a)^(x))=b,\quad a,b \gt 0\]
According to the “naive” algorithm that we used 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 that can already be solved. For example:
\[\begin(align)& ((2)^(x))=8\Rightarrow ((2)^(x))=((2)^(3))\Rightarrow x=3; \\& ((3)^(-x))=81\Rightarrow ((3)^(-x))=((3)^(4))\Rightarrow -x=4\Rightarrow x=-4; \\& ((5)^(2x))=125\Rightarrow ((5)^(2x))=((5)^(3))\Rightarrow 2x=3\Rightarrow x=\frac(3)( 2). \\\end(align)\]
And oddly enough, this scheme works in about 90% of cases. What then about the remaining 10%? The remaining 10% are slightly “schizophrenic” exponential equations of the form:
\[((2)^(x))=3;\quad ((5)^(x))=15;\quad ((4)^(2x))=11\]
Well, to what power do you need to raise 2 to get 3? First? But no: $((2)^(1))=2$ is not enough. Second? No either: $((2)^(2))=4$ is too much. Which one then?
Knowledgeable students have probably already guessed: in such cases, when it is not possible to solve it “beautifully”, the “heavy artillery” - logarithms - comes into play. 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: this formula (it is also the main 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 look at our equation and this formula:
\[\begin(align)& ((2)^(x))=3 \\& a=((b)^(((\log )_(b))a)) \\\end(align) \]
If we assume that $a=3$ is our original number on the right, and $b=2$ is the very base exponential function, to which we want to reduce the right-hand side, we get the following:
\[\begin(align)& a=((b)^(((\log )_(b))a))\Rightarrow 3=((2)^(((\log )_(2))3 )); \\& ((2)^(x))=3\Rightarrow ((2)^(x))=((2)^(((\log )_(2))3))\Rightarrow x=( (\log )_(2))3. \\\end(align)\]
We received a slightly strange answer: $x=((\log )_(2))3$. In some other task, many would have doubts with such an answer and would begin to double-check their solution: what if an error had crept in somewhere? I hasten to please you: there is no error here, and logarithms in the roots of exponential equations are a completely typical situation. So get used to it. :)
Now let’s solve the remaining two equations by analogy:
\[\begin(align)& ((5)^(x))=15\Rightarrow ((5)^(x))=((5)^(((\log )_(5))15)) \Rightarrow x=((\log )_(5))15; \\& ((4)^(2x))=11\Rightarrow ((4)^(2x))=((4)^(((\log )_(4))11))\Rightarrow 2x=( (\log )_(4))11\Rightarrow x=\frac(1)(2)((\log )_(4))11. \\\end(align)\]
That's all! By the way, the last answer can be written differently:
We introduced a factor into the argument of the logarithm. But no one is stopping us from adding this factor to 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 to decide.
Thus, we have learned to solve any exponential equations of the form $((a)^(x))=b$, where the numbers $a$ and $b$ are strictly positive. However, the harsh reality of our world is that such simple tasks you will meet very, very rarely. More often than not you will come across something like this:
\[\begin(align)& ((4)^(x))+((4)^(x-1))=((4)^(x+1))-11; \\& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2.7)^(1-x))=0.09. \\\end(align)\]
So how can we solve this? Can this be solved at all? And if so, how?
Don't panic. All these equations quickly and easily reduce to the simple formulas that we have already considered. You just need to remember a couple of tricks from the algebra course. And of course, there are no 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 complex it may be, one way or another must be reduced to the simplest equations - the 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: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
- Do some weird shit. Or even some crap called "convert an equation";
- At the output, get the simplest expressions of the form $((4)^(x))=4$ or something else like that. Moreover, one initial equation can give several such expressions at once.
Everything is clear with the first point - even my cat can write the equation on a piece of paper. The third point also seems to be more or less clear - we have already solved a whole bunch of such equations above.
But what about the second point? What kind of transformations? Convert what into what? And How?
Well, let's find out. First of all, I would like to note the following. All exponential equations are divided into two types:
- The equation is composed of exponential functions with the same base. Example: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
- The formula contains exponential functions with different bases. Examples: $((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x))$ and $((100)^(x-1) )\cdot ((2,7)^(1-x))=$0.09.
Let's start with equations of the first type - they are the easiest to solve. And in solving them, we will be helped by such a technique as highlighting stable expressions.
Isolating a stable expression
Let's look at this equation again:
\[((4)^(x))+((4)^(x-1))=((4)^(x+1))-11\]
What do we see? The four are raised to different degrees. But all these powers are simple sums of the variable $x$ with other numbers. Therefore, it is necessary to remember the rules for working with degrees:
\[\begin(align)& ((a)^(x+y))=((a)^(x))\cdot ((a)^(y)); \\& ((a)^(x-y))=((a)^(x)):((a)^(y))=\frac(((a)^(x)))(((a )^(y))). \\\end(align)\]
Simply put, addition 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 degrees from our equation:
\[\begin(align)& ((4)^(x-1))=\frac(((4)^(x)))(((4)^(1)))=((4)^ (x))\cdot \frac(1)(4); \\& ((4)^(x+1))=((4)^(x))\cdot ((4)^(1))=((4)^(x))\cdot 4. \ \\end(align)\]
Let's rewrite the original equation taking this fact into account, and then collect all the terms on the left:
\[\begin(align)& ((4)^(x))+((4)^(x))\cdot \frac(1)(4)=((4)^(x))\cdot 4 -eleven; \\& ((4)^(x))+((4)^(x))\cdot \frac(1)(4)-((4)^(x))\cdot 4+11=0. \\\end(align)\]
The first four terms contain the element $((4)^(x))$ - let’s take it out of the bracket:
\[\begin(align)& ((4)^(x))\cdot \left(1+\frac(1)(4)-4 \right)+11=0; \\& ((4)^(x))\cdot \frac(4+1-16)(4)+11=0; \\& ((4)^(x))\cdot \left(-\frac(11)(4) \right)=-11. \\\end(align)\]
It remains to divide both sides of the equation by the fraction $-\frac(11)(4)$, i.e. essentially multiply by the inverted fraction - $-\frac(4)(11)$. We get:
\[\begin(align)& ((4)^(x))\cdot \left(-\frac(11)(4) \right)\cdot \left(-\frac(4)(11) \right )=-11\cdot \left(-\frac(4)(11) \right); \\& ((4)^(x))=4; \\& ((4)^(x))=((4)^(1)); \\& x=1. \\\end(align)\]
That's all! We have reduced the original equation to its simplest form and obtained the final answer.
At the same time, in the process of solving we discovered (and even took it out of the bracket) the common factor $((4)^(x))$ - this is a stable expression. It can be designated as a new variable, or you can simply express it carefully and get the answer. In any case, the key principle of the solution is as follows:
Find in the original equation a stable expression containing a variable that is easily distinguished from all exponential functions.
The good news is that almost every exponential equation allows you to isolate such a stable expression.
But the bad news is that these expressions can be quite tricky and can be quite difficult to identify. So let's look at one more problem:
\[((5)^(x+2))+((0,2)^(-x-1))+4\cdot ((5)^(x+1))=2\]
Perhaps someone now has a question: “Pasha, are you stoned? There are different bases here – 5 and 0.2.” But let's try converting the power to base 0.2. For example, let’s get rid of the decimal fraction by reducing it to a regular one:
\[((0,2)^(-x-1))=((0,2)^(-\left(x+1 \right)))=((\left(\frac(2)(10 ) \right))^(-\left(x+1 \right)))=((\left(\frac(1)(5) \right))^(-\left(x+1 \right)) )\]
As you can see, the number 5 still appeared, albeit in the denominator. At the same time, the indicator was rewritten as negative. Now let’s remember one of the most important rules for working with degrees:
\[((a)^(-n))=\frac(1)(((a)^(n)))\Rightarrow ((\left(\frac(1)(5) \right))^( -\left(x+1 \right)))=((\left(\frac(5)(1) \right))^(x+1))=((5)^(x+1))\ ]
Here, of course, I was lying a little. Because for a complete understanding, the formula for getting rid of negative indicators had to be written like this:
\[((a)^(-n))=\frac(1)(((a)^(n)))=((\left(\frac(1)(a) \right))^(n ))\Rightarrow ((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(\frac(5)(1) \ right))^(x+1))=((5)^(x+1))\]
On the other hand, nothing prevented us from working with just fractions:
\[((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(((5)^(-1)) \ right))^(-\left(x+1 \right)))=((5)^(\left(-1 \right)\cdot \left(-\left(x+1 \right) \right) ))=((5)^(x+1))\]
But in this case, you need to be able to raise a power to another power (let me remind you: in this case, the indicators are added together). But I didn’t have to “reverse” the fractions - perhaps this will be easier for some. :)
In any case, the original exponential equation will be rewritten as:
\[\begin(align)& ((5)^(x+2))+((5)^(x+1))+4\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+5\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(1))\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(x+2))=2; \\& 2\cdot ((5)^(x+2))=2; \\& ((5)^(x+2))=1. \\\end(align)\]
So it turns out that the original equation can be solved even more simply than the one previously considered: here you don’t even need to select a stable expression - everything has been reduced by itself. It remains only to remember that $1=((5)^(0))$, from which we get:
\[\begin(align)& ((5)^(x+2))=((5)^(0)); \\& x+2=0; \\& x=-2. \\\end(align)\]
That's the solution! We got the final answer: $x=-2$. At the same time, I would like to note one technique that greatly simplified all calculations for us:
In exponential equations, be sure to get rid of decimals, convert them to regular ones. This will allow you to see the same bases of degrees and greatly simplify the solution.
Let us now move on to more complex equations in which there are different bases that cannot be reduced to each other using powers at all.
Using the Degrees Property
Let me remind you that we have two more particularly harsh equations:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2.7)^(1-x))=0.09. \\\end(align)\]
The main difficulty here is that it is not clear what to give and to what basis. Where set expressions? Where are the same grounds? There is none of this.
But let's try to go a different way. If there are no ready-made identical bases, you can try to find them by factoring the existing bases.
Let's start with the first equation:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& 21=7\cdot 3\Rightarrow ((21)^(3x))=((\left(7\cdot 3 \right))^(3x))=((7)^(3x))\ cdot((3)^(3x)). \\\end(align)\]
But you can do the opposite - make the number 21 from the numbers 7 and 3. This is especially easy to do on the left, since the indicators of both degrees are the same:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((\left(7\cdot 3 \right))^(x+ 6))=((21)^(x+6)); \\& ((21)^(x+6))=((21)^(3x)); \\& x+6=3x; \\& 2x=6; \\& x=3. \\\end(align)\]
That's all! You took the exponent outside the product and immediately got a beautiful equation that can be solved in a couple of lines.
Now let's look at the second equation. Everything is much more complicated here:
\[((100)^(x-1))\cdot ((2.7)^(1-x))=0.09\]
\[((100)^(x-1))\cdot ((\left(\frac(27)(10) \right))^(1-x))=\frac(9)(100)\]
In this case, the fractions turned out to be irreducible, but if something could be reduced, be sure to reduce it. Often, interesting reasons will appear with which you can already work.
Unfortunately, nothing special appeared for us. But we see that the exponents on the left in the product are opposite:
Let me remind you: to get rid of the minus sign in the indicator, you just need to “flip” the fraction. Well, let's rewrite the original equation:
\[\begin(align)& ((100)^(x-1))\cdot ((\left(\frac(10)(27) \right))^(x-1))=\frac(9 )(100); \\& ((\left(100\cdot \frac(10)(27) \right))^(x-1))=\frac(9)(100); \\& ((\left(\frac(1000)(27) \right))^(x-1))=\frac(9)(100). \\\end(align)\]
In the second line, we simply took the total exponent out of the product from the bracket according to the rule $((a)^(x))\cdot ((b)^(x))=((\left(a\cdot b \right))^ (x))$, and in the last one they simply multiplied the number 100 by a fraction.
Now note that the numbers on the left (at the base) and on the right are somewhat similar. How? Yes, it’s obvious: they are powers of the same number! We have:
\[\begin(align)& \frac(1000)(27)=\frac(((10)^(3)))(((3)^(3)))=((\left(\frac( 10)(3) \right))^(3)); \\& \frac(9)(100)=\frac(((3)^(2)))(((10)^(3)))=((\left(\frac(3)(10) \right))^(2)). \\\end(align)\]
Thus, our equation will be rewritten as follows:
\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(3 )(10)\right))^(2))\]
\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(10 )(3) \right))^(3\left(x-1 \right)))=((\left(\frac(10)(3) \right))^(3x-3))\]
In this case, on the right you can also get a degree with the same base, for which it is enough to simply “turn over” the fraction:
\[((\left(\frac(3)(10) \right))^(2))=((\left(\frac(10)(3) \right))^(-2))\]
Our equation will finally take the form:
\[\begin(align)& ((\left(\frac(10)(3) \right))^(3x-3))=((\left(\frac(10)(3) \right)) ^(-2)); \\& 3x-3=-2; \\& 3x=1; \\& x=\frac(1)(3). \\\end(align)\]
That's the solution. Its main idea boils down to the fact that even with on different grounds we are trying, by hook or by crook, to reduce these bases to the same thing. They help us with this elementary transformations equations and rules for working with powers.
But what rules and when to use? How do you understand that in one equation you need to divide both sides by something, and in another you need to factor the base of the exponential function?
The answer to this question will come with experience. Try your hand at first simple equations, and then gradually complicate the tasks - and very soon your skills will be enough to solve any exponential equation from the same Unified State Exam or any independent/test work.
And to help you in this difficult matter, I suggest downloading a set of equations for independent decision. 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 simple training will not be enough either. :)