Examples:

\(4^x=32\)
\(5^(2x-1)-5^(2x-3)=4.8\)
\((\sqrt(7))^(2x+2)-50\cdot(\sqrt(7))^(x)+7=0\)

How to Solve Exponential Equations

When solving any exponential equation, we strive to bring it to the form \(a^(f(x))=a^(g(x))\), and then make the transition to equality of exponents, that is:

\(a^(f(x))=a^(g(x))\) \(⇔\) \(f(x)=g(x)\)

For example:\(2^(x+1)=2^2\) \(⇔\) \(x+1=2\)

Important! From the same logic, two requirements for such a transition follow:
- number in left and right should be the same;
- the degrees on the left and right must be “pure”, that is, there should be no multiplication, division, etc.

For example:

To reduce the equation to the form \(a^(f(x))=a^(g(x))\) and are used.

Example . Solve the exponential equation \(\sqrt(27)·3^(x-1)=((\frac(1)(3)))^(2x)\)
Solution:

\(\sqrt(27)·3^(x-1)=((\frac(1)(3)))^(2x)\)		We know that \(27 = 3^3\). Taking this into account, we transform the equation.
\(\sqrt(3^3)·3^(x-1)=((\frac(1)(3)))^(2x)\)		By the property of the root \(\sqrt[n](a)=a^(\frac(1)(n))\) we obtain that \(\sqrt(3^3)=((3^3))^( \frac(1)(2))\). Next, using the property of degree \((a^b)^c=a^(bc)\), we obtain \(((3^3))^(\frac(1)(2))=3^(3 \ cdot \frac(1)(2))=3^(\frac(3)(2))\).
\(3^(\frac(3)(2))\cdot 3^(x-1)=(\frac(1)(3))^(2x)\)		We also know that \(a^b·a^c=a^(b+c)\). Applying this to the left side, we get: \(3^(\frac(3)(2))·3^(x-1)=3^(\frac(3)(2)+ x-1)=3^ (1.5 + x-1)=3^(x+0.5)\).
\(3^(x+0.5)=(\frac(1)(3))^(2x)\)		Now remember that: \(a^(-n)=\frac(1)(a^n)\). This formula can also be used in the opposite direction: \(\frac(1)(a^n) =a^(-n)\). Then \(\frac(1)(3)=\frac(1)(3^1) =3^(-1)\).
\(3^(x+0.5)=(3^(-1))^(2x)\)		Applying the property \((a^b)^c=a^(bc)\) to the right side, we obtain: \((3^(-1))^(2x)=3^((-1) 2x) =3^(-2x)\).
\(3^(x+0.5)=3^(-2x)\)		And now our bases are equal and there are no interfering coefficients, etc. So we can make the transition.
		Example . Solve the exponential equation \(4^(x+0.5)-5 2^x+2=0\) Solution: \(4^(x+0.5)-5 2^x+2=0\) We again use the power property \(a^b \cdot a^c=a^(b+c)\) in the opposite direction. \(4^x 4^(0.5)-5 2^x+2=0\) Now remember that \(4=2^2\). \((2^2)^x·(2^2)^(0.5)-5·2^x+2=0\) Using the properties of degrees, we transform: \((2^2)^x=2^(2x)=2^(x 2)=(2^x)^2\) \((2^2)^(0.5)=2^(2 0.5)=2^1=2.\) \(2·(2^x)^2-5·2^x+2=0\) We look carefully at the equation and see that the replacement \(t=2^x\) suggests itself. \(t_1=2\) \(t_2=\frac(1)(2)\) However, we found the values ​​of \(t\), and we need \(x\). We return to the X's, making a reverse replacement. \(2^x=2\) \(2^x=\frac(1)(2)\) Let's transform the second equation using the negative power property... \(2^x=2^1\) \(2^x=2^(-1)\) ...and we decide until the answer. \(x_1=1\) \(x_2=-1\) Answer : \(-1; 1\). The question remains - how to understand when to use which method? This comes with experience. Until you have worked it out, use the general recommendation to solve complex tasks- “If you don’t know what to do, do what you can.” That is, look for how you can transform the equation in principle, and try to do it - what if what happens? The main thing is to make only mathematically based transformations. Exponential equations without solutions Let's look at two more situations that often confuse students: - a positive number to the power is equal to zero, for example, \(2^x=0\); - a positive number is equal to a power of a negative number, for example, \(2^x=-4\). Let's try to solve by brute force. If x is a positive number, then as x grows, the entire power \(2^x\) will only increase: \(x=1\); \(2^1=2\) \(x=2\); \(2^2=4\) \(x=3\); \(2^3=8\). \(x=0\); \(2^0=1\) Also by. Negative X's remain. Remembering the property \(a^(-n)=\frac(1)(a^n)\), we check: \(x=-1\); \(2^(-1)=\frac(1)(2^1) =\frac(1)(2)\) \(x=-2\); \(2^(-2)=\frac(1)(2^2) =\frac(1)(4)\) \(x=-3\); \(2^(-3)=\frac(1)(2^3) =\frac(1)(8)\) Despite the fact that the number becomes smaller with each step, it will never reach zero. So the negative degree did not save us. We come to a logical conclusion: A positive number to any degree will remain a positive number. Thus, both equations above have no solutions. Exponential equations with different bases In practice, sometimes we encounter exponential equations with different bases that are not reducible to each other, and at the same time with the same exponents. They look like this: \(a^(f(x))=b^(f(x))\), where \(a\) and \(b\) are positive numbers. For example: \(7^(x)=11^(x)\) \(5^(x+2)=3^(x+2)\) \(15^(2x-1)=(\frac(1)(7))^(2x-1)\) Such equations can easily be solved by dividing by any of the sides of the equation (usually divided by the right side, that is, by \(b^(f(x))\). You can divide this way because a positive number is positive to any power (that is, we do not divide by zero). \(\frac(a^(f(x)))(b^(f(x)))\) \(=1\) Example . Solve the exponential equation \(5^(x+7)=3^(x+7)\) Solution: \(5^(x+7)=3^(x+7)\) Here we won’t be able to turn a five into a three, or vice versa (at least without using ). This means we cannot come to the form \(a^(f(x))=a^(g(x))\). However, the indicators are the same. Let's divide the equation by the right side, that is, by \(3^(x+7)\) (we can do this because we know that three will not be zero to any degree). \(\frac(5^(x+7))(3^(x+7))\) \(=\)\(\frac(3^(x+7))(3^(x+7) )\) Now remember the property \((\frac(a)(b))^c=\frac(a^c)(b^c)\) and use it on the left in the opposite direction. On the right, we simply reduce the fraction. \((\frac(5)(3))^(x+7)\) \(=1\) It would seem that things didn't get any better. But remember one more property of power: \(a^0=1\), in other words: “any number to the zero power is equal to \(1\).” The converse is also true: “one can be represented as any number to the zero power.” Let's take advantage of this by making the base on the right the same as on the left. \((\frac(5)(3))^(x+7)\) \(=\) \((\frac(5)(3))^0\) Voila! Let's get rid of the bases. We are writing a response. Answer : \(-7\). Sometimes the “sameness” of exponents is not obvious, but skillful use of the properties of exponents resolves this issue. Example . Solve the exponential equation \(7^( 2x-4)=(\frac(1)(3))^(-x+2)\) Solution: \(7^( 2x-4)=(\frac(1)(3))^(-x+2)\) The equation looks very sad... Not only can the bases not be reduced to the same number (seven will in no way be equal to \(\frac(1)(3)\)), but also the exponents are different... However, let's use the left exponent deuce. \(7^( 2(x-2))=(\frac(1)(3))^(-x+2)\) Remembering the property \((a^b)^c=a^(b·c)\) , we transform from the left: \(7^(2(x-2))=7^(2·(x-2))=(7^2)^(x-2)=49^(x-2)\). \(49^(x-2)=(\frac(1)(3))^(-x+2)\) Now, remembering the property of negative power \(a^(-n)=\frac(1)(a)^n\), we transform from the right: \((\frac(1)(3))^(-x+2) =(3^(-1))^(-x+2)=3^(-1(-x+2))=3^(x-2)\) \(49^(x-2)=3^(x-2)\) Hallelujah! The indicators are the same! Acting according to the scheme already familiar to us, we solve before the answer. Answer : \(2\).

Exponential equations. As you know, the Unified State Examination includes simple equations. We have already considered some - these are logarithmic, trigonometric, rational. Here are the exponential equations.

In a recent article we worked with exponential expressions, it will be useful. The equations themselves are solved simply and quickly. You just need to know the properties of exponents and... About thisfurther.

Let us list the properties of exponents:

The zero power of any number is equal to one.

A corollary from this property:

A little more theory.

An exponential equation is an equation containing a variable in the exponent, that is, it is an equation of the form:

f(x) expression that contains a variable

Solution methods exponential equations

1. As a result of transformations, the equation can be reduced to the form:

Then we apply the property:

2. Upon obtaining an equation of the form a f (x) = b using the definition of logarithm, we get:

3. As a result of transformations, you can obtain an equation of the form:

Logarithm applied:

Express and find x.

In tasks Unified State Exam options It will be enough to use the first method.

That is, it is necessary to represent the left and right sides in the form of powers with the same base, and then we equate the exponents and solve the usual linear equation.

Consider the equations:

Find the root of equation 4 1–2x = 64.

It is necessary to ensure that the left and right sides contain exponential expressions with the same base. We can represent 64 as 4 to the power of 3. We get:

4 1–2x = 4 3

1 – 2x = 3

– 2x = 2

x = – 1

Examination:

4 1–2 (–1) = 64

4 1 + 2 = 64

4 3 = 64

64 = 64

Answer: –1

Find the root of equation 3 x–18 = 1/9.

It is known that

So 3 x-18 = 3 -2

The bases are equal, we can equate the indicators:

x – 18 = – 2

x = 16

Examination:

3 16–18 = 1/9

3 –2 = 1/9

1/9 = 1/9

Answer: 16

Find the root of the equation:

Let's represent the fraction 1/64 as one-fourth to the third power:

2x – 19 = 3

2x = 22

x = 11

Examination:

Answer: 11

Find the root of the equation:

Let's imagine 1/3 as 3 –1, and 9 as 3 squared, we get:

(3 –1) 8–2x = 3 2

3 –1∙(8–2x) = 3 2

3 –8+2x = 3 2

Now we can equate the indicators:

– 8+2x = 2

2x = 10

x = 5

Examination:

Answer: 5

26654. Find the root of the equation:

Solution:

Answer: 8.75

Indeed, no matter what power we raise a positive number a to, we cannot get a negative number.

Any exponential equation after appropriate transformations is reduced to solving one or more simple ones.In this section we will also look at solving some equations, don’t miss it!That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

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:

Start solution given equation costs from the 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.

This is the name for equations of the form where the unknown is in both the exponent and the base of the power.

You can specify a completely clear algorithm for solving an equation of the form. To do this, you need to pay attention to the fact that when Oh) not equal to zero, one and minus one equality of powers with on the same grounds(be it positive or negative) is possible only if the exponents are equal. That is, all the roots of the equation will be the roots of the equation f(x) = g(x) The converse statement is not true, when Oh)< 0 and fractional values f(x) And g(x) expressions Oh) f(x) And

Oh) g(x) lose their meaning. That is, when moving from to f(x) = g(x)(for and extraneous roots may appear, which need to be excluded by checking against the original equation. And cases a = 0, a = 1, a = -1 need to be considered separately.

So, to completely solve the equation, we consider the cases:

a(x) = O f(x) And g(x) will be positive numbers, then this is the solution. Otherwise, no

a(x) = 1. The roots of this equation are also the roots of the original equation.

a(x) = -1. If, for a value of x that satisfies this equation, f(x) And g(x) are integers of the same parity (either both even or both odd), then this is the solution. Otherwise, no

When and we solve the equation f(x)= g(x) and by substituting the obtained results into the original equation we cut off the extraneous roots.

Examples of solving exponential-power equations.

Example No. 1.

1) x - 3 = 0, x = 3. because 3 > 0, and 3 2 > 0, then x 1 = 3 is the solution.

2) x - 3 = 1, x 2 = 4.

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

4) x - 3 ? 0 and x? ± 1. x = x 2, x = 0 or x = 1. For x = 0, (-3) 0 = (-3) 0 - this solution is correct: x 4 = 0. For x = 1, (-2) 1 = (-2) 1 - this solution is correct x 5 = 1.

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

Example No. 2.

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

1) x - 1 = 0 or x = 1, = 0, 0 0 is not a solution.

2) x - 1 = 1 x 1 = 2.

3) x - 1 = -1 x 2 = 0 does not fit in ODZ.

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

1º. Exponential equations are called equations containing a variable in an exponent.

Solving exponential equations is based on the property of powers: two powers with the same base are equal if and only if their exponents are equal.

2º. Basic methods for solving exponential equations:

1) the simplest equation has a solution;

2) an equation of the form logarithmic to the base a reduce to form ;

3) an equation of the form is equivalent to the equation ;

4) equation of the form is equivalent to the equation.

5) an equation of the form is reduced through substitution to an equation, and then a set of simple exponential equations is solved;

6) equation with reciprocal reciprocals by substitution they reduce to an equation, and then solve a set of equations;

7) equations homogeneous with respect to a g(x) And b g(x) given that kind through replacement they are reduced to an equation, and then a set of equations is solved.

Classification of exponential equations.

1. Equations solved by going to one base.

Example 18. Solve the equation .

Solution: Let's take advantage of the fact that all bases of powers are powers of the number 5: .

2. Equations solved by passing to one exponent.

These equations are solved by transforming the original equation to the form , which is reduced to its simplest using the property of proportion.

Example 19. Solve the equation:

3. Equations solved by taking the common factor out of brackets.

If each exponent in an equation differs from the other by a certain number, then the equations are solved by putting the exponent with the smallest exponent out of brackets.

Example 20. Solve the equation.

Solution: Let’s take the degree with the smallest exponent out of brackets on the left side of the equation:

Example 21. Solve the equation

Solution: Let's group separately on the left side of the equation the terms containing powers with base 4, on the right side - with base 3, then put the powers with the smallest exponent out of brackets:

4. Equations that reduce to quadratic (or cubic) equations.

The following equations are reduced to a quadratic equation for the new variable y:

a) the type of substitution, in this case;

b) the type of substitution , and .

Example 22. Solve the equation .

Solution: Let's make a change of variable and solve quadratic equation:

.

Answer: 0; 1.

5. Equations that are homogeneous with respect to exponential functions.

An equation of the form is a homogeneous equation of the second degree with respect to the unknowns a x And b x. Such equations are reduced by first dividing both sides by and then substituting them into quadratic equations.

Example 23. Solve the equation.

Solution: Divide both sides of the equation by:

Putting , we get a quadratic equation with roots .

Now the problem comes down to solving a set of equations . From the first equation we find that . The second equation has no roots, since for any value x.

Answer: -1/2.

6. Rational equations with respect to exponential functions.

Example 24. Solve the equation.

Solution: Divide the numerator and denominator of the fraction by 3 x and instead of two we get one exponential function:

7. Equations of the form .

Such equations with a set of admissible values ​​(APV), determined by the condition, by taking the logarithm of both sides of the equation are reduced to an equivalent equation, which in turn are equivalent to a set of two equations or.

Example 25. Solve the equation: .

.

Didactic material.

Solve the equations:

1. ; 2. ; 3. ;

4. ; 5. ; 6. ;

9. ; 10. ; 11. ;

14. ; 15. ;

16. ; 17. ;

18. ; 19. ;

20. ; 21. ;

22. ; 23. ;

24. ; 25. .

26. Find the product of the roots of the equation .

27. Find the sum of the roots of the equation .

Find the meaning of the expression:

28. , where x 0– root of the equation;

29. , where x 0 – whole root equations .

Solve the equation:

31. ; 32. .

Answers: 1. 0; 2. -2/9; 3. 1/36; 4. 0, 0.5; 5.0; 6.0; 7. -2; 8.2; 9. 1, 3; 10. 8; 11.5; 12.1; 13. ¼; 14.2; 15. -2, -1; 16. -2, 1; 17.0; 18.1; 19.0; 20. -1, 0; 21. -2, 2; 22. -2, 2; 23.4; 24. -1, 2; 25. -2, -1, 3; 26. -0.3; 27.3; 28.11; 29.54; 30. -1, 0, 2, 3; 31. ; 32. .

Topic No. 8.

Exponential inequalities.

1º. An inequality containing a variable in the exponent is called exponential inequality.

2º. The solution to exponential inequalities of the form is based on the following statements:

if , then the inequality is equivalent to ;

if , then the inequality is equivalent to .

When solving exponential inequalities, use the same techniques as when solving exponential equations.

Example 26. Solve inequality (method of transition to one base).

Solution: Because , then the given inequality can be written as: . Since , then this inequality is equivalent to the inequality .

Solving the last inequality, we get .

Example 27. Solve the inequality: ( by taking the common factor out of brackets).

Solution: Let's take out brackets on the left side of the inequality, on the right side of the inequality and divide both sides of the inequality by (-2), changing the sign of the inequality to the opposite:

Since , then when moving to inequality of indicators, the sign of inequality again changes to the opposite. We get. Thus, the set of all solutions to this inequality is the interval.

Example 28. Solve inequality ( by introducing a new variable).

Solution: Let . Then this inequality will take the form: or , whose solution is the interval .

From here. Since the function increases, then .

Didactic material.

Specify the set of solutions to the inequality:

1. ; 2. ; 3. ;

6. At what values x Do the points on the function graph lie below the straight line?

7. At what values x Do the points on the graph of the function lie at least as low as the straight line?

Solve the inequality:

8. ; 9. ; 10. ;

13. Specify the largest integer solution to the inequality .

14. Find the product of the largest integer and the smallest integer solutions to the inequality .

Solve the inequality:

15. ; 16. ; 17. ;

18. ; 19. ; 20. ;

21. ; 22. ; 23. ;

24. ; 25. ; 26. .

Find the domain of the function:

27. ; 28. .

29. Find the set of argument values ​​for which the values ​​of each of the functions are greater than 3:

And .

Answers: 11.3; 12.3; 13. -3; 14.1; 15. (0; 0.5); 16. ; 17. (-1; 0)U(3; 4); 18. [-2; 2]; 19. (0; +∞); 20. (0; 1); 21. (3; +∞); 22. (-∞; 0)U(0.5; +∞); 23. (0; 1); 24. (-1; 1); 25. (0; 2]; 26. (3; 3.5)U (4; +∞); 27. (-∞; 3)U(5); 28. )