How to solve exponential equations with different bases. Solution of exponential equations. Examples
Lecture: "Methods for solving exponential equations."
1 . exponential equations.
Equations containing unknowns in the exponent are called exponential equations. The simplest of these is the equation ax = b, where a > 0 and a ≠ 1.
1) For b< 0 и b = 0 это уравнение, согласно свойству 1 показательной функции, не имеет решения.
2) For b > 0, using the monotonicity of the function and the root theorem, the equation has a single root. In order to find it, b must be represented as b = aс, ax = bс ó x = c or x = logab.
The exponential equations, through algebraic transformations, lead to standard equations, which are solved using the following methods:
1) method of reduction to one base;
2) evaluation method;
3) graphic method;
4) the method of introducing new variables;
5) factorization method;
6) exponential - power equations;
7) exponential with a parameter.
2 . Method of reduction to one basis.
The method is based on the following property of degrees: if two degrees are equal and their bases are equal, then their exponents are equal, i.e., the equation must be tried to be reduced to the form
Examples. Solve the equation:
1 . 3x=81;
Let's represent the right side of the equation in the form 81 = 34 and write the equation equivalent to the original 3 x = 34; x = 4. Answer: 4.
2. https://pandia.ru/text/80/142/images/image004_8.png" width="52" height="49"> and go to the equation for exponents 3x+1 = 3 – 5x; 8x = 4; x = 0.5 Answer: 0.5
3. https://pandia.ru/text/80/142/images/image006_8.png" width="105" height="47">
Note that the numbers 0.2, 0.04, √5, and 25 are powers of 5. Let's take advantage of this and transform the original equation as follows:
,
whence 5-x-1 = 5-2x-2 ó - x - 1 = - 2x - 2, from which we find the solution x = -1. Answer: -1.
5. 3x = 5. By definition of the logarithm, x = log35. Answer: log35.
6. 62x+4 = 33x. 2x+8.
Let's rewrite the equation as 32x+4.22x+4 = 32x.2x+8, i.e..png" width="181" height="49 src="> Hence x - 4 =0, x = 4. Answer: 4.
7 . 2∙3x+1 - 6∙3x-2 - 3x = 9. Using the properties of powers, we write the equation in the form e. x+1 = 2, x =1. Answer: 1.
Bank of tasks No. 1.
Solve the equation:
Test number 1.
1) 0 2) 4 3) -2 4) -4 |
|
A2 32x-8 = √3. | 1)17/4 2) 17 3) 13/2 4) -17/4 |
A3 | 1) 3;1 2) -3;-1 3) 0;2 4) no roots |
1) 7;1 2) no roots 3) -7;1 4) -1;-7 |
|
A5 | 1) 0;2; 2) 0;2;3 3) 0 4) -2;-3;0 |
A6 | 1) -1 2) 0 3) 2 4) 1 |
Test #2
A1 | 1) 3 2) -1;3 3) -1;-3 4) 3;-1 |
A2 | 1) 14/3 2) -14/3 3) -17 4) 11 |
A3 | 1) 2;-1 2) no roots 3) 0 4) -2;1 |
A4 | 1) -4 2) 2 3) -2 4) -4;2 |
A5 | 1) 3 2) -3;1 3) -1 4) -1;3 |
3 Assessment method.
The root theorem: if the function f (x) increases (decreases) on the interval I, the number a is any value taken by f on this interval, then the equation f (x) = a has a single root on the interval I.
When solving equations by the estimation method, this theorem and the monotonicity properties of the function are used.
Examples. Solve Equations: 1. 4x = 5 - x.
Decision. Let's rewrite the equation as 4x + x = 5.
1. if x \u003d 1, then 41 + 1 \u003d 5, 5 \u003d 5 is true, then 1 is the root of the equation.
The function f(x) = 4x is increasing on R and g(x) = x is increasing on R => h(x)= f(x)+g(x) is increasing on R as the sum of increasing functions, so x = 1 is the only root of the equation 4x = 5 – x. Answer: 1.
2.
Decision. We rewrite the equation in the form 
.
1. if x = -1, then 
, 3 = 3-true, so x = -1 is the root of the equation.
2. prove that it is unique.
3. The function f(x) = - decreases on R, and g(x) = - x - decreases on R => h(x) = f(x) + g(x) - decreases on R, as the sum of decreasing functions . So by the root theorem, x = -1 is the only root of the equation. Answer: -1.
Bank of tasks No. 2. solve the equation
a) 4x + 1 = 6 - x;
b) 
c) 2x – 2 =1 – x;
4. Method for introducing new variables.
The method is described in section 2.1. The introduction of a new variable (substitution) is usually carried out after transformations (simplification) of the terms of the equation. Consider examples.
Examples.
R eat equation: 1.
.
Let's rewrite the equation differently: https://pandia.ru/text/80/142/images/image030_0.png" width="128" height="48 src="> i.e..png" width="210" height ="45">
Decision. Let's rewrite the equation differently: 
Denote https://pandia.ru/text/80/142/images/image035_0.png" width="245" height="57"> - not suitable.
t = 4 => https://pandia.ru/text/80/142/images/image037_0.png" width="268" height="51"> is an irrational equation. Note that 
The solution to the equation is x = 2.5 ≤ 4, so 2.5 is the root of the equation. Answer: 2.5.
Decision. Let's rewrite the equation in the form and divide both sides by 56x+6 ≠ 0. We get the equation
2x2-6x-7 = 2x2-6x-8 +1 = 2(x2-3x-4)+1, so..png" width="118" height="56">
The roots of the quadratic equation - t1 = 1 and t2<0, т. е..png" width="200" height="24">.
Decision . We rewrite the equation in the form
and note that it is a homogeneous equation of the second degree.
Divide the equation by 42x, we get 
Replace https://pandia.ru/text/80/142/images/image049_0.png" width="16" height="41 src="> .
Answer: 0; 0.5.
Task Bank #3. solve the equation
b) 
G) 
Test #3 with a choice of answers. Minimum level.
A1 | 1) -0.2;2 2) log52 3) –log52 4) 2 |
А2 0.52x – 3 0.5x +2 = 0. | 1) 2;1 2) -1;0 3) no roots 4) 0 |
1) 0 2) 1; -1/3 3) 1 4) 5 |
|
A4 52x-5x - 600 = 0. | 1) -24;25 2) -24,5; 25,5 3) 25 4) 2 |
1) no roots 2) 2;4 3) 3 4) -1;2 |
Test #4 with a choice of answers. General level.
A1 | 1) 2;1 2) ½;0 3)2;0 4) 0 |
А2 2x – (0.5)2x – (0.5)x + 1 = 0 | 1) -1;1 2) 0 3) -1;0;1 4) 1 |
1) 64 2) -14 3) 3 4) 8 |
|
1)-1 2) 1 3) -1;1 4) 0 |
|
A5 | 1) 0 2) 1 3) 0;1 4) no roots |
5. Method of factorization.
1. Solve the equation: 5x+1 - 5x-1 = 24.
Solution..png" width="169" height="69"> , from where
2. 6x + 6x+1 = 2x + 2x+1 + 2x+2.
Decision. Let us take out 6x on the left side of the equation, and 2x on the right side. We get the equation 6x(1+6) = 2x(1+2+4) ó 6x = 2x.
Since 2x >0 for all x, we can divide both sides of this equation by 2x without fear of losing solutions. We get 3x = 1ó x = 0.
3. 
Decision. We solve the equation by factoring.
We select the square of the binomial 
4. https://pandia.ru/text/80/142/images/image067_0.png" width="500" height="181">
x = -2 is the root of the equation.
Equation x + 1 = 0 " style="border-collapse:collapse;border:none">
A1 5x-1 +5x -5x+1 = -19.
1) 1 2) 95/4 3) 0 4) -1
A2 3x+1 +3x-1 =270.
1) 2 2) -4 3) 0 4) 4
A3 32x + 32x+1 -108 = 0. x=1.5
1) 0,2 2) 1,5 3) -1,5 4) 3
1) 1 2) -3 3) -1 4) 0
A5 2x -2x-4 = 15.x=4
1) -4 2) 4 3) -4;4 4) 2
Test #6 General level.
A1 (22x-1)(24x+22x+1)=7. | 1) ½ 2) 2 3) -1;3 4) 0.2 |
A2 | 1) 2.5 2) 3;4 3) log43/2 4) 0 |
A3 2x-1-3x=3x-1-2x+2. | 1) 2 2) -1 3) 3 4) -3 |
A4 | 1) 1,5 2) 3 3) 1 4) -4 |
A5 | 1) 2 2) -2 3) 5 4) 0 |
6. Exponential - power equations.
The exponential equations are adjoined by the so-called exponential-power equations, i.e. equations of the form (f(x))g(x) = (f(x))h(x).
If it is known that f(x)>0 and f(x) ≠ 1, then the equation, like the exponential one, is solved by equating the exponents g(x) = f(x).
If the condition does not exclude the possibility of f(x)=0 and f(x)=1, then we have to consider these cases when solving the exponential power equation.
1..png" width="182" height="116 src=">
2. 
Decision. x2 +2x-8 - makes sense for any x, because a polynomial, so the equation is equivalent to the set
https://pandia.ru/text/80/142/images/image078_0.png" width="137" height="35">
b) 
7. Exponential equations with parameters.
1. For what values of the parameter p does the equation 4 (5 – 3)2 +4p2–3p = 0 (1) have a unique solution?
Decision. Let us introduce the change 2x = t, t > 0, then equation (1) will take the form t2 – (5p – 3)t + 4p2 – 3p = 0. (2)
The discriminant of equation (2) is D = (5p – 3)2 – 4(4p2 – 3p) = 9(p – 1)2.
Equation (1) has a unique solution if equation (2) has one positive root. This is possible in the following cases.
1. If D = 0, that is, p = 1, then equation (2) will take the form t2 – 2t + 1 = 0, hence t = 1, therefore, equation (1) has a unique solution x = 0.
2. If p1, then 9(p – 1)2 > 0, then equation (2) has two different roots t1 = p, t2 = 4p – 3. The set of systems satisfies the condition of the problem
Substituting t1 and t2 into the systems, we have
https://pandia.ru/text/80/142/images/image084_0.png" alt="(!LANG:no35_11" width="375" height="54"> в зависимости от параметра a?!}
Decision. Let be 
then equation (3) will take the form t2 – 6t – a = 0. (4)
Let us find the values of the parameter a for which at least one root of equation (4) satisfies the condition t > 0.
Let us introduce the function f(t) = t2 – 6t – a. The following cases are possible.
https://pandia.ru/text/80/142/images/image087.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_14.gif" align="left" width="215" height="73 src=">где t0 - абсцисса вершины параболы и D - дискриминант квадратного трехчлена f(t);!}
https://pandia.ru/text/80/142/images/image089.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_16.gif" align="left" width="60" height="51 src=">!}
Case 2. Equation (4) has a unique positive solution if
D = 0, if a = – 9, then equation (4) will take the form (t – 3)2 = 0, t = 3, x = – 1.
Case 3. Equation (4) has two roots, but one of them does not satisfy the inequality t > 0. This is possible if
https://pandia.ru/text/80/142/images/image092.png" alt="(!LANG:no35_17" width="267" height="63">!}
Thus, at a 0 equation (4) has a single positive root 
. Then equation (3) has a unique solution 
For a< – 9 уравнение (3) корней не имеет.
if a< – 9, то корней нет; если – 9 < a < 0, то
if a = – 9, then x = – 1;
if a 0, then
Let us compare the methods for solving equations (1) and (3). Note that when solving equation (1) was reduced to a quadratic equation, the discriminant of which is a full square; thus, the roots of equation (2) were immediately calculated by the formula of the roots of the quadratic equation, and then conclusions were drawn regarding these roots. Equation (3) was reduced to a quadratic equation (4), the discriminant of which is not a perfect square, therefore, when solving equation (3), it is advisable to use theorems on the location of the roots of a square trinomial and a graphical model. Note that equation (4) can be solved using the Vieta theorem.
Let's solve more complex equations.
Task 3. Solve the equation 
Decision. ODZ: x1, x2.
Let's introduce a replacement. Let 2x = t, t > 0, then, as a result of transformations, the equation will take the form t2 + 2t – 13 – a = 0. (*) Find the values of a for which at least one root of the equation (*) satisfies the condition t > 0.
https://pandia.ru/text/80/142/images/image098.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_23.gif" align="left" width="71" height="68 src=">где t0 - абсцисса вершины f(t) = t2 + 2t – 13 – a, D - дискриминант квадратного трехчлена f(t).!}
https://pandia.ru/text/80/142/images/image100.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_25.gif" align="left" width="360" height="32 src=">!}
https://pandia.ru/text/80/142/images/image102.png" alt="(!LANG:http://1september.ru/ru/mat/2002/35/no35_27.gif" align="left" width="218" height="42 src=">!}
Answer: if a > - 13, a 11, a 5, then if a - 13,
a = 11, a = 5, then there are no roots.
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2. Guzeev technology: from reception to philosophy.
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Mathematics at school No. 2, 1987, pp. 9 - 11.
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Mathematics at School No. 6, 1990, p. 37-40.
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Mathematics at School No. 1, 1997, p. 32-36.
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Mathematics at School No. 1, 1993, p. 27 - 28.
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Mathematics at School No. 2, 1989, p. ten.
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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: $56x-11=0$; $((x)^(2))+5x+4=0$; $((x)^(2))-12x+32=0$ etc. To be able to solve such constructions is absolutely necessary in order not to “hang” in the topic that will be discussed now.
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 complicated to you, some of them, on the contrary, are too simple. But all of them are united by one important feature: they contain an exponential function $f\left(x \right)=((a)^(x))$. Thus, we introduce the definition:
An exponential equation is any equation that contains an exponential function, i.e. an 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. Understood the definition. Now the question is: how to solve all this crap? The answer is both simple and complex at the same time.
Let's start with the good news: from my experience with many students, I can say that for most of them, exponential equations are much easier than the same logarithms, and even more so trigonometry.
But there is also bad news: sometimes the compilers of problems for all kinds of textbooks and exams are visited by “inspiration”, and their drug-inflamed brain begins to produce such brutal equations that it becomes problematic not only for students to solve them - 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 the number 2 be raised to get the number 4? Perhaps the second? After all, $((2)^(2))=2\cdot 2=4$ — and we have obtained 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 is a little more difficult. 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 exponents (similar to the formula $((a)^(-n))= \frac(1)(((a)^(n)))$).
Finally, only a select few guess that these facts can be combined and the output is the following result:
\[\frac(1)(25)=\frac(1)(((5)^(2)))=((5)^(-2))\]
Thus, our original equation will be rewritten as follows:
\[((5)^(2x-3))=\frac(1)(25)\Rightarrow ((5)^(2x-3))=((5)^(-2))\]
And now this is already completely solved! On the left side of the equation there is an exponential function, on the right side of the equation there is an exponential function, there is nothing but them anywhere else. Therefore, it is possible to “discard” the bases and stupidly equate the indicators:
We got the simplest linear equation that any student can solve in just a couple of lines. Okay, in four lines:
\[\begin(align)& 2x-3=-2 \\& 2x=3-2 \\& 2x=1 \\& x=\frac(1)(2) \\\end(align)\]
If you 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 assimilation of this topic, it is too early for you to take on exponential equations.
\[((9)^(x))=-3\]
Well, how do you decide? First thought: $9=3\cdot 3=((3)^(2))$, so the original equation can be rewritten like this:
\[((\left(((3)^(2)) \right))^(x))=-3\]
Then we recall that when raising a degree to a power, the indicators are multiplied:
\[((\left(((3)^(2)) \right))^(x))=((3)^(2x))\Rightarrow ((3)^(2x))=-(( 3)^(1))\]
\[\begin(align)& 2x=-1 \\& x=-\frac(1)(2) \\\end(align)\]
And for such a decision, we get an honestly deserved deuce. For we, with the equanimity of a Pokémon, sent the minus sign in front of the three to the power of this very three. And you can't do that. And that's why. Take a look at the different powers of the triple:
\[\begin(matrix) ((3)^(1))=3& ((3)^(-1))=\frac(1)(3)& ((3)^(\frac(1)( 2)))=\sqrt(3) \\ ((3)^(2))=9& ((3)^(-2))=\frac(1)(9)& ((3)^(\ frac(1)(3)))=\sqrt(3) \\ ((3)^(3))=27& ((3)^(-3))=\frac(1)(27)& (( 3)^(-\frac(1)(2)))=\frac(1)(\sqrt(3)) \\\end(matrix)\]
Compiling this tablet, I didn’t pervert as soon as I did: I considered positive degrees, and negative, and even fractional ones ... well, where is at least one negative number here? He is not! And it cannot be, because the exponential function $y=((a)^(x))$, firstly, always takes only positive values (no matter how much you multiply one or divide 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 $((9)^(x))=-3$? No, there are no roots. And in this sense, exponential equations are very similar to quadratic ones - there may also be no roots. But if in quadratic equations the number of roots is determined by the discriminant (the discriminant is positive - 2 roots, negative - no roots), then in exponential equations it all 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 $b>0$. 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 write down that there are no roots.
This knowledge will help us many times over when we have to solve more complex problems. In the meantime, enough lyrics - it's time to study the basic algorithm for solving exponential equations.
How to solve exponential equations
So, let's formulate the problem. It is necessary to solve the exponential equation:
\[((a)^(x))=b,\quad a,b>0\]
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, which 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 about the other 10% then? The remaining 10% are slightly "schizophrenic" exponential equations of the form:
\[((2)^(x))=3;\quad ((5)^(x))=15;\quad ((4)^(2x))=11\]
To what power do you need to raise 2 to get 3? In the first? But no: $((2)^(1))=2$ is not enough. In the second? Neither: $((2)^(2))=4$ is too much. What then?
Knowledgeable students have probably already guessed: in such cases, when it is impossible to solve “beautifully”, “heavy artillery” is connected to the case - logarithms. Let me remind you that using logarithms, any positive number can be represented as a power of any other positive number (with the exception of 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 the logarithm) will haunt you for a very long time and “emerge” 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 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 got a slightly strange answer: $x=((\log )_(2))3$. In some other task, with such an answer, many would doubt and begin to double-check their solution: what if there was a mistake somewhere? I hasten to please you: there is no error here, and logarithms in the roots of exponential equations are quite a typical situation. So get used to it. :)
Now we solve by analogy the remaining two equations:
\[\begin(align)& ((5)^(x))=15\Rightarrow ((5)^(x))=((5)^(((\log )_(5))15)) \Rightarrow x=((\log )_(5))15; \\& ((4)^(2x))=11\Rightarrow ((4)^(2x))=((4)^(((\log )_(4))11))\Rightarrow 2x=( (\log )_(4))11\Rightarrow x=\frac(1)(2)((\log )_(4))11. \\\end(align)\]
That's all! By the way, the last answer can be written differently:
It was we who introduced the multiplier into the argument of the logarithm. But no one prevents 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 decision is up to you.
Thus, we have learned to solve any exponential equations of the form $((a)^(x))=b$, where the numbers $a$ and $b$ are strictly positive. However, the harsh reality of our world is that such simple tasks will meet you very, very rarely. More often you will come across something like this:
\[\begin(align)& ((4)^(x))+((4)^(x-1))=((4)^(x+1))-11; \\& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2,7)^(1-x))=0.09. \\\end(align)\]
Well, how do you decide? Can this be resolved at all? And if so, how?
No panic. All these equations are quickly and simply reduced to those simple formulas that we have already considered. You just need to know to remember a couple of tricks from the algebra course. And of course, there are no rules for working with degrees here. I'll talk about all this now. :)
Transformation of exponential equations
The first thing to remember is that any exponential equation, no matter how complex it may be, one way or another must be reduced to the simplest equations - the very 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 stupid shit. Or even some crap called "transform the equation";
- At the output, get the simplest expressions like $((4)^(x))=4$ or something else like that. Moreover, one initial equation can give several such expressions at once.
With the first point, everything is clear - even my cat can write the equation on a leaf. With the third point, too, it seems, it is more or less clear - we have already solved a whole bunch of such equations above.
But what about the second point? What are the transformations? 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: $((4)^(x))+((4)^(x-1))=((4)^(x+1))-11$;
- 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 their solution we will be helped by such a technique as the selection of stable expressions.
Highlighting 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 of exponents can be converted to a product of powers, and subtraction is easily converted to division. Let's try to apply these formulas to the powers from our equation:
\[\begin(align)& ((4)^(x-1))=\frac(((4)^(x)))(((4)^(1)))=((4)^ (x))\cdot \frac(1)(4); \\& ((4)^(x+1))=((4)^(x))\cdot ((4)^(1))=((4)^(x))\cdot 4. \ \\end(align)\]
We rewrite the original equation taking into account this fact, and then we 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 parts 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 reduced the original equation to the simplest and got the final answer.
At the same time, in the process of solving, we discovered (and even took out of the bracket) the common factor $((4)^(x))$ - this is the stable expression. It can be designated as a new variable, or you can simply accurately express it and get an 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 admits such a stable expression.
But there is also bad news: such expressions can be very tricky, and it can be quite difficult to distinguish them. So let's look at another 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? Here are different bases - 5 and 0.2. But let's try to convert a power with base 0.2. For example, let's get rid of the decimal fraction, 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 we recall 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 cheated a little. Because for a complete understanding, the formula for getting rid of negative indicators had to be written as follows:
\[((a)^(-n))=\frac(1)(((a)^(n)))=((\left(\frac(1)(a) \right))^(n ))\Rightarrow ((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(\frac(5)(1) \ right))^(x+1))=((5)^(x+1))\]
On the other hand, nothing prevented us from working with only one fraction:
\[((\left(\frac(1)(5) \right))^(-\left(x+1 \right)))=((\left(((5)^(-1)) \ right))^(-\left(x+1 \right)))=((5)^(\left(-1 \right)\cdot \left(-\left(x+1 \right) \right) ))=((5)^(x+1))\]
But in this case, you need to be able to raise a degree to another degree (I remind you: in this case, the indicators are added up). But I didn’t have to “flip” the fractions - perhaps for someone it will be easier. :)
In any case, the original exponential equation will be rewritten as:
\[\begin(align)& ((5)^(x+2))+((5)^(x+1))+4\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+5\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(1))\cdot ((5)^(x+1))=2; \\& ((5)^(x+2))+((5)^(x+2))=2; \\& 2\cdot ((5)^(x+2))=2; \\& ((5)^(x+2))=1. \\\end(align)\]
So it turns out that the original equation is even easier to solve than the previously considered one: here you don’t even need to single out a stable expression - everything has been reduced by itself. It remains only to remember that $1=((5)^(0))$, whence we get:
\[\begin(align)& ((5)^(x+2))=((5)^(0)); \\&x+2=0; \\&x=-2. \\\end(align)\]
That's the whole solution! We got the final answer: $x=-2$. At the same time, I would like to note one trick that greatly simplified all the calculations for us:
In exponential equations, be sure to get rid of decimal fractions, translate them into ordinary ones. This will allow you to see the same bases of the degrees and greatly simplify the solution.
Now let's move on to more complex equations in which there are different bases, which are generally not reduced to each other with the help of powers.
Using the exponent property
Let me remind you that we have two more particularly harsh equations:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& ((100)^(x-1))\cdot ((2,7)^(1-x))=0.09. \\\end(align)\]
The main difficulty here is that it is not clear what and to what basis to lead. Where are the fixed expressions? Where are the common 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 available bases.
Let's start with the first equation:
\[\begin(align)& ((7)^(x+6))\cdot ((3)^(x+6))=((21)^(3x)); \\& 21=7\cdot 3\Rightarrow ((21)^(3x))=((\left(7\cdot 3 \right))^(3x))=((7)^(3x))\ cdot ((3)^(3x)). \\\end(align)\]
But you can do the opposite - make up the number 21 from the numbers 7 and 3. It is especially easy to do this 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 out of 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. Here everything is much more complicated:
\[((100)^(x-1))\cdot ((2,7)^(1-x))=0.09\]
\[((100)^(x-1))\cdot ((\left(\frac(27)(10) \right))^(1-x))=\frac(9)(100)\]
In this case, the fractions turned out to be irreducible, but if something could be reduced, be sure to reduce it. This will often result in interesting grounds that you can already work with.
Unfortunately, we haven't come up with anything. 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 exponent, you just need to “flip” the fraction. So 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 bracketed the total from the product according to the rule $((a)^(x))\cdot ((b)^(x))=((\left(a\cdot b \right))^ (x))$, and in the latter they simply multiplied the number 100 by a fraction.
Now note that the numbers on the left (at the base) and on the right are somewhat similar. How? Yes, obviously: 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:
\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(3 )(10) \right))^(2))\]
\[((\left(((\left(\frac(10)(3) \right))^(3)) \right))^(x-1))=((\left(\frac(10 )(3) \right))^(3\left(x-1 \right)))=((\left(\frac(10)(3) \right))^(3x-3))\]
At the same time, on the right, you can also get a degree with the same base, for which it is enough just to “flip” the fraction:
\[((\left(\frac(3)(10) \right))^(2))=((\left(\frac(10)(3) \right))^(-2))\]
Finally, our equation will take the form:
\[\begin(align)& ((\left(\frac(10)(3) \right))^(3x-3))=((\left(\frac(10)(3) \right)) ^(-2)); \\& 3x-3=-2; \\& 3x=1; \\& x=\frac(1)(3). \\\end(align)\]
That's the whole solution. Its main idea boils down to the fact that even with different reasons, we try by hook or by crook to reduce these reasons to the same one. In this we are helped by elementary transformations of equations and the rules for working with powers.
But what rules and when to use? How to understand that in one equation you need to divide both sides by something, and in another - to decompose the base of the exponential function into factors?
The answer to this question will come with experience. Try your hand at first on simple equations, and then gradually complicate the tasks - and very soon your skills will be enough to solve any exponential equation from the same USE or any independent / test work.
And to help you in this difficult task, I suggest downloading a set of equations on my website for an independent solution. All equations have answers, so you can always check yourself.
At the stage of preparation for the final testing, 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 carefully master the theory, memorize the formulas and understand the principle of solving such equations. Having learned to cope with this type of tasks, graduates will be able to count on high scores when passing the exam in mathematics.
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The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. Studying on our site, you will be able to identify gaps in knowledge and pay attention to precisely those tasks that cause the greatest difficulties.
Teachers of "Shkolkovo" collected, systematized and presented all the material necessary for the successful passing of the exam in the simplest and most accessible form.
The main definitions and formulas are presented in the "Theoretical Reference" section.
For a better assimilation of the material, we recommend that you practice the assignments. Carefully review the examples of exponential equations with solutions presented on this page in order to understand the calculation algorithm. After that, proceed with the tasks in the "Catalogs" 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.
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First, let's recall the basic formulas of degrees and their properties.
Product of a number a happens 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 \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 is always at the bottom, and the variable x degree or measure.
Let us give more examples of exponential equations.
2 x *5=10
16x-4x-6=0
Now let's look at how exponential equations are solved?
Let's take a simple equation:
2 x = 2 3
Such an example can be solved even in the mind. 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 this decision should be made:
2 x = 2 3
x = 3
To solve this equation, we removed same grounds(that is, deuces) and wrote down what was left, these are degrees. We got the answer we were looking for.
Now let's summarize our solution.
Algorithm for solving the exponential equation:
1. Need to check the same whether the bases of the equation on the right and on the 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 some examples:
Let's start 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 has turned out.
x=4 - 2
x=2
Answer: x=2
In the following example, you can see that the bases are different, these are 3 and 9.
3 3x - 9 x + 8 = 0
To begin with, we transfer 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 \u003d (3 2) x + 8
We get 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2 x + 16
3 3x \u003d 3 2x + 16 now it is clear that the bases on the left and right sides are the same and equal to three, which means we can discard them and equate the degrees.
3x=2x+16 got the simplest equation
3x-2x=16
x=16
Answer: x=16.
Let's look at the following example:
2 2x + 4 - 10 4 x \u003d 2 4
First of all, we look at the bases, the bases are different two and four. And we need to be the same. We transform the quadruple according to 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 interfere with us. 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 - 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 whole equation by 6:
Imagine 4=2 2:
2 2x \u003d 2 2 bases are the same, discard them and equate the degrees.
2x \u003d 2 turned out to be the simplest equation. We divide it by 2, we get
x = 1
Answer: x = 1.
Let's solve the equation:
9 x - 12*3 x +27= 0
Let's transform:
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, it is clear that the first triple has a degree twice (2x) than the second (just x). In this case, you can decide substitution method. The number with the smallest degree is replaced by:
Then 3 2x \u003d (3 x) 2 \u003d t 2
We replace all degrees with x's in the equation with t:
t 2 - 12t + 27 \u003d 0
We get a quadratic equation. We solve through the discriminant, we get:
D=144-108=36
t1 = 9
t2 = 3
Back to Variable x.
We take t 1:
t 1 \u003d 9 \u003d 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 \u003d 3 \u003d 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 \u003d 2; x 2 = 1.
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Solution of exponential equations. Examples.
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
What exponential equation? This is an equation in which the unknowns (x) and expressions with them are in indicators some degrees. And only there! It is important.
There you are examples of exponential equations:
3 x 2 x = 8 x + 3
Note! In the bases of degrees (below) - only numbers. AT indicators degrees (above) - a wide variety of expressions with x. If, suddenly, an x appears in the equation somewhere other than the indicator, for example:
this will be a mixed type equation. Such equations do not have clear rules for solving. We will not consider them for now. Here we will deal with solution of exponential equations in its purest form.
In fact, even pure exponential equations are not always clearly solved. But there are certain types of exponential equations that can and should be solved. These are the types we'll be looking at.
Solution of the simplest exponential equations.
Let's start with something very basic. For example:
Even without any theory, by simple selection it is clear that x = 2. Nothing more, right!? No other x value rolls. And now let's look at the solution of this tricky exponential equation:
What have we done? We, in fact, just threw out the same bottoms (triples). Completely thrown out. And, what pleases, hit the mark!
Indeed, if in the exponential equation on the left and on the right are the same numbers in any degree, these numbers can be removed and equal exponents. Mathematics allows. It remains to solve a much simpler equation. It's good, right?)
However, let's remember ironically: you can remove the bases only when the base numbers on the left and right are in splendid isolation! Without any neighbors and coefficients. Let's say in the equations:
2 x +2 x + 1 = 2 3 , or
You can't remove doubles!
Well, we have mastered the most important thing. How to move from evil exponential expressions to simpler equations.
"Here are those times!" - you say. "Who will give such a primitive on the control and exams!?"
Forced to agree. Nobody will. But now you know where to go when solving confusing examples. It is necessary to bring it to mind, when the same base number is on the left - on the right. Then everything will be easier. Actually, this is the classics of mathematics. We take the original example and transform it to the desired us mind. According to the rules of mathematics, of course.
Consider examples that require some additional effort to bring them to the simplest. Let's call them simple exponential equations.
Solution of simple exponential equations. Examples.
When solving exponential equations, the main rules are actions with powers. Without knowledge of these actions, nothing will work.
To actions with degrees, one must add personal observation and ingenuity. Do we need the same base numbers? So we are looking for them in the example in an explicit or encrypted form.
Let's see how this is done in practice?
Let's give us an example:
2 2x - 8 x+1 = 0
First glance at grounds. They... They are different! Two and eight. But it's too early to be discouraged. It's time to remember that
Two and eight are relatives in degree.) It is quite possible to write down:
8 x+1 = (2 3) x+1
If we recall the formula from actions with powers:
(a n) m = a nm ,
it generally works great:
8 x+1 = (2 3) x+1 = 2 3(x+1)
The original example looks like this:
2 2x - 2 3(x+1) = 0
We transfer 2 3 (x+1) to the right (no one canceled the elementary actions of mathematics!), we get:
2 2x \u003d 2 3 (x + 1)
That's practically all. Removing bases:
We solve this monster and get
This is the correct answer.
In this example, knowing the powers of two helped us out. We identified in the eight, the encrypted deuce. This technique (encoding common bases under different numbers) is a very popular trick in exponential equations! Yes, even in logarithms. One must be able to recognize the powers of other numbers in numbers. This is extremely important for solving exponential equations.
The fact is that raising any number to any power is not a problem. Multiply, even on a piece of paper, and that's all. For example, everyone can raise 3 to the fifth power. 243 will turn out if you know the multiplication table.) But in exponential equations, much more often it is necessary not to raise to a power, but vice versa ... what number to what extent hides behind the number 243, or, say, 343... No calculator will help you here.
You need to know the powers of some numbers by sight, yes ... Shall we practice?
Determine what powers and what numbers are numbers:
2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.
Answers (in a mess, of course!):
5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .
If you look closely, you can see a strange fact. There are more answers than questions! Well, it happens... For example, 2 6 , 4 3 , 8 2 is all 64.
Let's assume that you have taken note of the information about acquaintance with numbers.) Let me remind you that for solving exponential equations, we apply the whole stock of mathematical knowledge. Including from the lower-middle classes. You didn't go straight to high school, did you?
For example, when solving exponential equations, putting the common factor out of brackets very often helps (hello to grade 7!). Let's see an example:
3 2x+4 -11 9 x = 210
And again, the first look - on the grounds! The bases of the degrees are different ... Three and nine. And we want them to be the same. Well, in this case, the desire is quite feasible!) Because:
9 x = (3 2) x = 3 2x
According to the same rules for actions with degrees:
3 2x+4 = 3 2x 3 4
That's great, you can write:
3 2x 3 4 - 11 3 2x = 210
We gave an example for the same reasons. So, what is next!? Threes cannot be thrown out ... Dead end?
Not at all. Remembering the most universal and powerful decision rule all math tasks:
If you don't know what to do, do what you can!
You look, everything is formed).
What is in this exponential equation can do? Yes, the left side directly asks for parentheses! The common factor of 3 2x clearly hints at this. Let's try, and then we'll see:
3 2x (3 4 - 11) = 210
3 4 - 11 = 81 - 11 = 70
The example keeps getting better and better!
We recall that in order to eliminate bases, we need a pure degree, without any coefficients. The number 70 bothers us. So we divide both sides of the equation by 70, we get:
Op-pa! Everything has been fine!
This is the final answer.
It happens, however, that taxiing out on the same grounds is obtained, but their liquidation is not. This happens in exponential equations of another type. Let's get this type.
Change of variable in solving exponential equations. Examples.
Let's solve the equation:
4 x - 3 2 x +2 = 0
First - as usual. Let's move on to the base. To the deuce.
4 x = (2 2) x = 2 2x
We get the equation:
2 2x - 3 2 x +2 = 0
And here we'll hang. The previous tricks will not work, no matter how you turn it. We'll have to get from the arsenal of another powerful and versatile way. It's called variable substitution.
The essence of the method is surprisingly simple. Instead of one complex icon (in our case, 2 x), we write another, simpler one (for example, t). Such a seemingly meaningless replacement leads to amazing results!) Everything just becomes clear and understandable!
So let
Then 2 2x \u003d 2 x2 \u003d (2 x) 2 \u003d t 2
We replace in our equation all powers with x's by t:
Well, it dawns?) Haven't forgotten quadratic equations yet? We solve through the discriminant, we get:
Here, the main thing is not to stop, as it happens ... This is not the answer yet, we need x, not t. We return to Xs, i.e. making a replacement. First for t 1:
That is,
One root was found. We are looking for the second one, from t 2:
Um... Left 2 x, Right 1... A hitch? Yes, not at all! It is enough to remember (from actions with degrees, yes ...) that a unity is any number to zero. Any. Whatever you need, we will put it. We need a two. Means:
Now that's all. Got 2 roots:
This is the answer.
At solving exponential equations at the end, some awkward expression is sometimes obtained. Type:
From the seven, a deuce through a simple degree does not work. They are not relatives ... How can I be here? Someone may be confused ... But the person who read on this site the topic "What is a logarithm?" , only smile sparingly and write down with a firm hand the absolutely correct answer:
There can be no such answer in tasks "B" on the exam. There is a specific number required. But in tasks "C" - easily.
This lesson provides examples of solving the most common exponential equations. Let's highlight the main one.
Practical Tips:
1. First of all, we look at grounds degrees. Let's see if they can't be done the same. Let's try to do this by actively using actions with powers. Do not forget that numbers without x can also be turned into powers!
2. We try to bring the exponential equation to the form when the left and right are the same numbers to any degree. We use actions with powers and factorization. What can be counted in numbers - we count.
3. If the second advice did not work, we try to apply the variable substitution. The result can be an equation that is easily solved. Most often - square. Or fractional, which also reduces to a square.
4. To successfully solve exponential equations, you need to know the degrees of some numbers "by sight".
As usual, at the end of the lesson you are invited to solve a little.) On your own. From simple to complex.
Solve exponential equations:
More difficult:
2 x + 3 - 2 x + 2 - 2 x \u003d 48
9 x - 8 3 x = 9
2 x - 2 0.5 x + 1 - 8 = 0
Find product of roots:
2 3-x + 2 x = 9
Happened?
Well, then the most complicated example (it is solved, however, in the mind ...):
7 0.13x + 13 0.7x+1 + 2 0.5x+1 = -3
What is more interesting? Then here's a bad example for you. Quite pulling on increased difficulty. I will hint that in this example, ingenuity and the most universal rule for solving all mathematical tasks saves.)
2 5x-1 3 3x-1 5 2x-1 = 720 x
An example is simpler, for relaxation):
9 2 x - 4 3 x = 0
And for dessert. Find the sum of the roots of the equation:
x 3 x - 9x + 7 3 x - 63 = 0
Yes Yes! This is a mixed type equation! Which we did not consider in this lesson. And what to consider them, they need to be solved!) This lesson is quite enough to solve the equation. Well, ingenuity is needed ... And yes, the seventh grade will help you (this is a hint!).
Answers (in disarray, separated by semicolons):
one; 2; 3; 4; there are no solutions; 2; -2; -5; 4; 0.
Is everything successful? Fine.
There is a problem? No problem! In Special Section 555, all these exponential equations are solved with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all sorts of exponential equations. Not only with these.)
One last fun question to consider. In this lesson, we worked with exponential equations. Why didn't I say a word about ODZ here? In equations, this is a very important thing, by the way ...
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By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.