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 specified 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 compilers 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 of negative powers (similar to 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 happened 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 the different powers of three:
\[\begin(matrix) ((3)^(1))=3& ((3)^(-1))=\frac(1)(3)& ((3)^(\frac(1)( 2)))=\sqrt(3) \\ ((3)^(2))=9& ((3)^(-2))=\frac(1)(9)& ((3)^(\ frac(1)(3)))=\sqrt(3) \\ ((3)^(3))=27& ((3)^(-3))=\frac(1)(27)& (( 3)^(-\frac(1)(2)))=\frac(1)(\sqrt(3)) \\\end(matrix)\]
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 exponential equations everything depends on what is to the right of the equal sign.
Thus, let us formulate the key conclusion: the simplest exponential equation of the form $((a)^(x))=b$ has a root if and only if $b>0$. 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:
\[((a)^(x))=b,\quad a,b>0\]
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 of the exponential function to which we so 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 multiplier to the argument of the logarithm. But no one is stopping us from adding this factor to the base:
Moreover, all three options are correct - it’s simple different shapes records of 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 can quickly and easily be reduced to simple formulas which 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 reasons. 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 will now have 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 decimal, bringing it to the usual:
\[((0,2)^(-x-1))=((0,2)^(-\left(x+1 \right)))=((\left(\frac(2)(10 ) \right))^(-\left(x+1 \right)))=((\left(\frac(1)(5) \right))^(-\left(x+1 \right)) )\]
As you can see, the number 5 still appeared, albeit in the denominator. At the same time, the indicator was rewritten as negative. And now let's remember one of the most important rules work 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 decimal fractions and convert them to ordinary 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 are the stable 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 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. His main idea boils down to the fact that even with different bases we try, by hook or by crook, to reduce these bases to the same thing. Elementary transformations of equations and rules for working with powers help us with this.
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 task, I suggest downloading a set of equations from my website for solving it yourself. All equations have answers, so you can always test yourself.
What is an exponential equation? Examples.
So, an exponential equation... A new unique exhibit in our general exhibition of a wide variety of equations!) As is almost always the case, the key word of any new mathematical term is the corresponding adjective that characterizes it. So it is here. The key word in the term “exponential equation” is the word "indicative". What does it mean? This word means that the unknown (x) is located in terms of any degrees. And only there! This is extremely important.
For example, these simple equations:
3 x +1 = 81
5 x + 5 x +2 = 130
4 2 2 x -17 2 x +4 = 0
Or even these monsters:
2 sin x = 0.5
Please immediately pay attention to one important thing: reasons degrees (bottom) – only numbers. But in indicators degrees (above) - a wide variety of expressions with an X. Absolutely any.) Everything depends on the specific equation. If, suddenly, x appears somewhere else in the equation, in addition to the indicator (say, 3 x = 18 + x 2), then such an equation will already be an equation mixed type. Such equations do not have clear rules for solving them. Therefore in this lesson we will not consider them. To the delight of the students.) Here we will consider only exponential equations in their “pure” form.
Generally speaking, not all and not always even pure exponential equations can be solved clearly. But among all the rich variety of exponential equations, there are certain types that can and should be solved. It is these types of equations that we will consider. And we’ll definitely solve the examples.) So let’s get comfortable and off we go! As in computer shooters, our journey will take place through levels.) From elementary to simple, from simple to intermediate and from intermediate to complex. Along the way, a secret level will also await you - techniques and methods for solving non-standard examples. Those that you won’t read about in most school textbooks... Well, and at the end, of course, a final boss awaits you in the form of homework.)
Level 0. What is the simplest exponential equation? Solving simple exponential equations.
First, let's look at some frank elementary stuff. You have to start somewhere, right? For example, this equation:
2 x = 2 2
Even without any theories, by simple logic and common sense it is clear that x = 2. There is no other way, right? No other meaning of X is suitable... And now let’s turn our attention to record of decision this cool exponential equation:
2 x = 2 2
X = 2
What happened to us? And the following happened. We actually took it and... simply threw out the same bases (twos)! Completely thrown out. And, the good news is, we hit the bull’s eye!
Yes, indeed, if in an exponential equation there are left and right the same numbers in any powers, then these numbers can be discarded and simply equate the exponents. Mathematics allows.) And then you can work separately with the indicators and solve a much simpler equation. Great, right?
Here is the key idea for solving any (yes, exactly any!) exponential equation: using identical transformations, it is necessary to ensure that the left and right sides of the equation are the same base numbers in various powers. And then you can safely remove the same bases and equate the exponents. And work with a simpler equation.
Now let’s remember the iron rule: it is possible to remove identical bases if and only if the base numbers on the left and right of the equation are in proud loneliness.
What does it mean, in splendid isolation? This means without any neighbors and coefficients. Let me explain.
For example, in Eq.
3 3 x-5 = 3 2 x +1
Threes cannot be removed! Why? Because on the left we have not just a lonely three to the degree, but work 3·3 x-5 . An extra three interferes: the coefficient, you understand.)
The same can be said about the equation
5 3 x = 5 2 x +5 x
Here, too, all the bases are the same - five. But on the right we don’t have a single power of five: there is a sum of powers!
In short, we have the right to remove identical bases only when our exponential equation looks like this and only like this:
af (x) = a g (x)
This type of exponential equation is called the simplest. Or, scientifically, canonical . And no matter what convoluted equation we have in front of us, we will, one way or another, reduce it to precisely this simplest (canonical) form. Or, in some cases, to totality equations of this type. Then our simplest equation can be written as general view rewrite it like this:
F(x) = g(x)
That's all. This would be an equivalent conversion. In this case, f(x) and g(x) can be absolutely any expressions with an x. Whatever.
Perhaps a particularly inquisitive student will wonder: why on earth do we so easily and simply discard the same bases on the left and right and equate the exponents? Intuition is intuition, but what if, in some equation and for some reason, this approach turns out to be incorrect? Is it always legal to throw out the same grounds? Unfortunately, for a rigorous mathematical answer to this interest Ask you need to dive quite deeply and seriously into the general theory of the structure and behavior of functions. And a little more specifically - in the phenomenon strict monotony. In particular, strict monotony exponential functiony= a x. Since it is the exponential function and its properties that underlie the solution of exponential equations, yes.) A detailed answer to this question will be given in a separate special lesson dedicated to solving complex non-standard equations using the monotonicity of different functions.)
Explaining this point in detail now would only blow the minds of the average schoolchild and scare him away ahead of time with a dry and heavy theory. I won’t do this.) Because our main task at the moment is learn to solve exponential equations! The simplest ones! Therefore, let’s not worry yet and boldly throw out the same reasons. This Can, take my word for it!) And then we solve the equivalent equation f(x) = g(x). As a rule, simpler than the original exponential.
It is assumed, of course, that at the moment people already know how to solve at least , and equations, without x’s in exponents.) For those who still don’t know how, feel free to close this page, follow the relevant links and fill in the old gaps. Otherwise you will have a hard time, yes...
I'm not talking about irrational, trigonometric and other brutal equations that can also emerge in the process of eliminating the foundations. But don’t be alarmed, we won’t consider outright cruelty in terms of degrees for now: it’s too early. We will train only on the simplest equations.)
Now let's look at equations that require some additional effort to reduce them to the simplest. For the sake of distinction, let's call them simple exponential equations. So, let's move to the next level!
Level 1. Simple exponential equations. Let's recognize the degrees! Natural indicators.
The key rules in solving any exponential equations are rules for dealing with degrees. Without this knowledge and skills, nothing will work. Alas. So, if there are problems with the degrees, then first you are welcome. In addition, we will also need . These transformations (two of them!) are the basis for solving all mathematical equations in general. And not only demonstrative ones. So, whoever forgot, also take a look at the link: I don’t just put them there.
But operations with powers and identity transformations alone are not enough. Personal observation and ingenuity are also required. We need the same reasons, don't we? So we examine the example and look for them in an explicit or disguised form!
For example, this equation:
3 2 x – 27 x +2 = 0
First look at grounds. They are different! Three and twenty seven. But it’s too early to panic and despair. It's time to remember that
27 = 3 3
Numbers 3 and 27 are relatives by degree! And close ones.) Therefore, we have every right to write:
27 x +2 = (3 3) x+2
Now let’s connect our knowledge about actions with degrees(and I warned you!). There is a very useful formula there:
(a m) n = a mn
If you now put it into action, it works out great:
27 x +2 = (3 3) x+2 = 3 3(x +2)
The original example now looks like this:
3 2 x – 3 3(x +2) = 0
Great, the bases of the degrees have leveled out. That's what we wanted. Half the battle is done.) Now we launch the basic identity transformation - move 3 3(x +2) to the right. No one has canceled the elementary operations of mathematics, yes.) We get:
3 2 x = 3 3(x +2)
What does this type of equation give us? And the fact that now our equation is reduced to canonical form: on the left and right there are the same numbers (threes) in powers. Moreover, both three are in splendid isolation. Feel free to remove the triples and get:
2x = 3(x+2)
We solve this and get:
X = -6
That's it. This is the correct answer.)
Now let’s think about the solution. What saved us in this example? Knowledge of the powers of three saved us. How exactly? We identified number 27 contains an encrypted three! This trick (encryption of the same base under different numbers) is one of the most popular in exponential equations! Unless it's the most popular. Yes, and in the same way, by the way. This is why observation and the ability to recognize powers of other numbers in numbers are so important in exponential equations!
Practical advice:
You need to know the powers of popular numbers. In face!
Of course, anyone can raise two to the seventh power or three to the fifth power. Not in my mind, but at least in a draft. But in exponential equations, much more often it is not necessary to raise to a power, but, on the contrary, to find out what number and to what power is hidden behind the number, say, 128 or 243. And this is more complicated than simple raising, you will agree. Feel the difference, as they say!
Since the ability to recognize degrees in person will be useful not only at this level, but also at the next ones, here’s a small task for you:
Determine what powers and what numbers the numbers are:
4; 8; 16; 27; 32; 36; 49; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729; 1024.
Answers (randomly, of course):
27 2 ; 2 10 ; 3 6 ; 7 2 ; 2 6 ; 9 2 ; 3 4 ; 4 3 ; 10 2 ; 2 5 ; 3 5 ; 7 3 ; 16 2 ; 2 7 ; 5 3 ; 2 8 ; 6 2 ; 3 3 ; 2 9 ; 2 4 ; 2 2 ; 4 5 ; 25 2 ; 4 4 ; 6 3 ; 8 2 ; 9 3 .
Yes Yes! Don't be surprised that there are more answers than tasks. For example, 2 8, 4 4 and 16 2 are all 256.
Level 2. Simple exponential equations. Let's recognize the degrees! Negative and fractional indicators.
At this level we are already using our knowledge of degrees to the fullest. Namely, we involve negative and fractional indicators in this fascinating process! Yes Yes! We need to increase our power, right?
For example, this terrible equation:
/image003.png){kind=small}
Again, the first glance is at the foundations. The reasons are different! And this time they are not even remotely similar to each other! 5 and 0.04... And to eliminate the bases, the same ones are needed... What to do?
It's OK! In fact, everything is the same, it’s just that the connection between the five and 0.04 is visually poorly visible. How can we get out? Let's move on to the number 0.04 as an ordinary fraction! And then, you see, everything will work out.)
0,04 = 4/100 = 1/25
Wow! It turns out that 0.04 is 1/25! Well, who would have thought!)
So how? Is it now easier to see the connection between the numbers 5 and 1/25? That's it...
And now according to the rules of actions with degrees with negative indicator You can write with a steady hand:
/image004.png){kind=small}
That is great. So we got to the same base - five. Now we replace the inconvenient number 0.04 in the equation with 5 -2 and get:
/image005.png){kind=small}
Again, according to the rules of operations with degrees, we can now write:
(5 -2) x -1 = 5 -2(x -1)
Just in case, I remind you (in case anyone doesn’t know) that basic rules actions with powers are valid for any indicators! Including for negative ones.) So, feel free to take and multiply the indicators (-2) and (x-1) according to the appropriate rule. Our equation gets better and better:
/image006.png){kind=small}
All! Apart from lonely fives, there is nothing else in the powers on the left and right. The equation is reduced to canonical form. And then - along the knurled track. We remove the fives and equate the indicators:
x 2 –6 x+5=-2(x-1)
The example is almost solved. All that's left is elementary middle school math - open (correctly!) the brackets and collect everything on the left:
x 2 –6 x+5 = -2 x+2
x 2 –4 x+3 = 0
We solve this and get two roots:
x 1 = 1; x 2 = 3
That's all.)
Now let's think again. IN in this example we again had to recognize the same number to different degrees! Namely, to see an encrypted five in the number 0.04. And this time - in negative degree! How did we do this? Right off the bat - no way. But after moving from the decimal fraction 0.04 to the common fraction 1/25, everything became clear! And then the whole decision went like clockwork.)
Therefore, another green practical advice.
If an exponential equation contains decimal fractions, then we move from decimal fractions to ordinary fractions. IN ordinary fractions It's much easier to recognize powers of many popular numbers! After recognition, we move from fractions to powers with negative exponents.
Keep in mind that this trick occurs very, very often in exponential equations! But the person is not in the subject. He looks, for example, at the numbers 32 and 0.125 and gets upset. Unbeknownst to him, this is one and the same two, only in different degrees... But you’re already in the know!)
Solve the equation:
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In! It looks like quiet horror... However, appearances are deceiving. This is the simplest exponential equation, despite its daunting appearance. And now I will show it to you.)
First, let’s look at all the numbers in the bases and coefficients. They are, of course, different, yes. But we will still take a risk and try to make them identical! Let's try to get to the same number in different powers. Moreover, preferably, the numbers are as small as possible. So, let's start decoding!
Well, with the four everything is immediately clear - it’s 2 2. So, that's something already.)
With a fraction of 0.25 – it’s still unclear. Need to check. Let's use practical advice - move from a decimal fraction to an ordinary fraction:
0,25 = 25/100 = 1/4
Much better already. Because now it is clearly visible that 1/4 is 2 -2. Great, and the number 0.25 is also akin to two.)
So far so good. But the worst number of all remains - square root of two! What to do with this pepper? Can it also be represented as a power of two? And who knows...
Well, let's dive into our treasury of knowledge about degrees again! This time we additionally connect our knowledge about roots. From the 9th grade course, you and I should have learned that any root, if desired, can always be turned into a degree with a fractional indicator.
Like this:
In our case:
/1.png){kind=small}
Wow! It turns out that the square root of two is 2 1/2. That's it!
That's fine! All our inconvenient numbers actually turned out to be an encrypted two.) I don’t argue, somewhere very sophisticatedly encrypted. But we are also improving our professionalism in solving such ciphers! And then everything is already obvious. In our equation we replace the numbers 4, 0.25 and the root of two by powers of two:
/image010.png)
All! The bases of all degrees in the example became the same - two. And now standard actions with degrees are used:
a ma n = a m + n
a m:a n = a m-n
(a m) n = a mn
For the left side you get:
2 -2 ·(2 2) 5 x -16 = 2 -2+2(5 x -16)
For the right side it will be:
/image011.png)
And now our evil equation looks like this:
/image012.png){kind=small}
For those who haven’t figured out exactly how this equation came about, then the question here is not about exponential equations. The question is about actions with degrees. I asked you to urgently repeat it to those who have problems!
Here is the finish line! Received canonical view exponential equation! So how? Have I convinced you that everything is not so scary? ;) We remove the twos and equate the indicators:
/image013.png){kind=small}
All that remains is to solve this linear equation. How? With the help of identical transformations, of course.) Decide what’s going on! Multiply both sides by two (to remove the fraction 3/2), move the terms with X's to the left, without X's to the right, bring similar ones, count - and you will be happy!
Everything should turn out beautifully:
X=4
Now let’s think about the solution again. In this example, we were helped by the transition from square root To degree with exponent 1/2. Moreover, only such a cunning transformation helped us reach everywhere same base(two), which saved the situation! And, if not for it, then we would have every chance to freeze forever and never cope with this example, yes...
Therefore, we do not neglect the next practical advice:
If an exponential equation contains roots, then we move from roots to powers with fractional exponents. Very often only such a transformation clarifies the further situation.
Of course, negative and fractional powers are already much more complex than natural powers. At least from the point of view visual perception and, especially, recognition from right to left!
It is clear that directly raising, for example, two to the power -3 or four to the power -3/2 is not so a big problem. For those in the know.)
But go, for example, immediately realize that
0,125 = 2 -3
Or
Here, only practice and rich experience rule, yes. And, of course, a clear idea, What is a negative and fractional degree? And - practical advice! Yes, yes, those same ones green.) I hope that they will still help you better navigate the entire diverse variety of degrees and significantly increase your chances of success! So let's not neglect them. I'm not in vain green I write sometimes.)
But if you get to know each other even with such exotic powers as negative and fractional ones, then your capabilities in solving exponential equations will expand enormously, and you will be able to handle almost any type of exponential equations. Well, if not any, then 80 percent of all exponential equations - for sure! Yes, yes, I'm not joking!
So, our first part of our introduction to exponential equations has come to its logical conclusion. And, as an intermediate workout, I traditionally suggest doing a little self-reflection.)
Exercise 1.
So that my words about deciphering negative and fractional powers do not go in vain, I suggest playing a little game!
Express numbers as powers of two:
Answers (in disarray):
Happened? Great! Then we do a combat mission - solve the simplest and simplest exponential equations!
Task 2.
Solve the equations (all answers are a mess!):
5 2x-8 = 25
2 5x-4 – 16 x+3 = 0
Answers:
x = 16
x 1 = -1; x 2 = 2
x = 5
Happened? Indeed, it’s much simpler!
Then we solve the next game:
/image018.png){kind=small}
(2 x +4) x -3 = 0.5 x 4 x -4
35 1-x = 0.2 - x ·7 x
Answers:
x 1 = -2; x 2 = 2
x = 0,5
x 1 = 3; x 2 = 5
And these examples are one left? Great! You are growing! Then here are some more examples for you to snack on:
/image019.png)
Answers:
x = 6
x = 13/31
x = -0,75
x 1 = 1; x 2 = 8/3
And is this decided? Well, respect! I take my hat off.) This means that the lesson was not in vain, and the initial level of solving exponential equations can be considered successfully mastered. Next levels and more are ahead complex equations! And new techniques and approaches. And non-standard examples. And new surprises.) All this is in the next lesson!
Did something go wrong? This means that most likely the problems are in . Or in . Or both at once. I'm powerless here. I can in Once again I can only suggest one thing - don’t be lazy and follow the links.)
To be continued.)
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Power or exponential equations are equations in which the variables are in powers and the base is a number. For example:
Solving an exponential equation comes down to 2 fairly simple steps:
1. You need to check whether the bases of the equation on the right and left are the same. If the reasons are not the same, we look for options to solve this example.
2. After the bases become the same, we equate the degrees and solve the resulting new equation.
Suppose we are given an exponential equation of the following form:
It is worth starting the solution of this equation with an analysis of the basis. The bases are different - 2 and 4, but to solve we need them to be the same, so we transform 4 using the following formula -\[ (a^n)^m = a^(nm):\]
We add to the original equation:
Let's take it out of brackets \
Let's express \
Since the degrees are the same, we discard them:
Answer: \
Where can I solve an exponential equation using an online solver?
You can solve the equation on our website https://site. A free online solver will allow you to solve the equation online any complexity in seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.
In this lesson we will look at solving more complex exponential equations and recall the basic theoretical principles regarding the exponential function.
1. Definition and properties of the exponential function, methods for solving the simplest exponential equations
Let us recall the definition and basic properties of the exponential function. The solution of all exponential equations and inequalities is based on these properties.
Exponential function is a function of the form , where the base is the degree and Here x is the independent variable, argument; y is the dependent variable, function.
Rice. 1. Graph of exponential function
The graph shows increasing and decreasing exponents, illustrating the exponential function with a base greater than one and less than one but greater than zero, respectively.
Both curves pass through the point (0;1)
Properties of the Exponential Function:
Domain: ;
Range of values: ;
The function is monotonic, increases with, decreases with.
A monotonic function takes each of its values given a single argument value.
When the argument increases from minus to plus infinity, the function increases from zero inclusive to plus infinity. On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero, not inclusive.
2. Solving standard exponential equations
Let us remind you how to solve the simplest exponential equations. Their solution is based on the monotonicity of the exponential function. Almost all complex exponential equations can be reduced to such equations.
The equality of exponents with equal bases is due to the property of the exponential function, namely its monotonicity.
Solution method:
Equalize the bases of degrees;
Equate the exponents.
Let's move on to consider more complex exponential equations; our goal is to reduce each of them to the simplest.
Let's get rid of the root on the left side and bring the degrees to the same base:
In order to reduce a complex exponential equation to its simplest, substitution of variables is often used.
Let's use the power property:
We are introducing a replacement. Let it be then 
Let's multiply the resulting equation by two and move all terms to the left side:
The first root does not satisfy the range of y values, so we discard it. We get:
Let's reduce the degrees to the same indicator:
Let's introduce a replacement:
Let it be then 
. With such a replacement, it is obvious that y takes on strictly positive values. We get:
We know how to solve such quadratic equations, we can write down the answer:
To make sure that the roots are found correctly, you can check using Vieta’s theorem, i.e., find the sum of the roots and their product and compare them with the corresponding coefficients of the equation.
We get:
3. Methodology for solving homogeneous exponential equations of the second degree
Let's study the following important type exponential equations:
Equations of this type are called homogeneous of the second degree with respect to the functions f and g. On the left side there is quadratic trinomial relative to f with parameter g or quadratic trinomial relative to g with parameter f.
Solution method:
This equation can be solved as a quadratic equation, but it is easier to do it differently. There are two cases to consider:
In the first case we get
In the second case, we have the right to divide by the highest degree and get:
It is necessary to introduce a change of variables, we obtain a quadratic equation for y:
Let us note that the functions f and g can be any, but we are interested in the case when these are exponential functions.
4. Examples of solving homogeneous equations
Let's move all the terms to the left side of the equation:
Since exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when:
We get:
Let's introduce a replacement: 
(according to the properties of the exponential function)
We got a quadratic equation:
We determine the roots using Vieta’s theorem:
The first root does not satisfy the range of values of y, we discard it, we get:
Let's use the properties of degrees and reduce all degrees to simple bases:
It's easy to notice the functions f and g:
Since exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when .