Solving exponential equations. Examples

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 have 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 developed it, use the general recommendation for solving complex problems - “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) We get:

\(\frac(a^(f(x)))(b^(f(x)))\) \(=1\)

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 from 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 degree \(a^(-n)=\frac(1)(a)^n\), 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 : \(2\).

Solving most mathematical problems in one way or another involves transforming numerical, algebraic or functional expressions. The above applies especially to the decision. In the versions of the Unified State Exam in mathematics, this type of problem includes, in particular, task C3. Learning to solve C3 tasks is important not only for the purpose of successfully passing the Unified State Exam, but also for the reason that this skill will be useful when studying a mathematics course in high school.

When completing C3 tasks, you have to solve various types of equations and inequalities. Among them are rational, irrational, exponential, logarithmic, trigonometric, containing modules (absolute values), as well as combined ones. This article discusses the main types of exponential equations and inequalities, as well as various methods for solving them. Read about solving other types of equations and inequalities in the “” section in articles devoted to methods for solving C3 problems from the Unified State Examination in mathematics.

Before we begin to analyze specific exponential equations and inequalities, as a math tutor, I suggest you brush up on some theoretical material that we will need.

Exponential function

What is an exponential function?

Function of the form y = a x, Where a> 0 and a≠ 1 is called exponential function.

Basic properties of exponential function y = a x:

Graph of an Exponential Function

The graph of the exponential function is exponent:

Graphs of exponential functions (exponents)

Solving exponential equations

Indicative are called equations in which the unknown variable is found only in exponents of some powers.

For solutions exponential equations you need to know and be able to use the following simple theorem:

Theorem 1. Exponential equation a f(x) = a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).

In addition, it is useful to remember the basic formulas and operations with degrees:

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Example 1. Solve the equation:

Solution: We use the above formulas and substitution:

The equation then becomes:

The discriminant of the resulting quadratic equation is positive:

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This means that this equation has two roots. We find them:

Moving on to reverse substitution, we get:

The second equation has no roots, since the exponential function is strictly positive throughout the entire domain of definition. Let's solve the second one:

Taking into account what was said in Theorem 1, we move on to the equivalent equation: x= 3. This will be the answer to the task.

Answer: x = 3.

Example 2. Solve the equation:

Solution: The equation has no restrictions on the range of permissible values, since the radical expression makes sense for any value x(exponential function y = 9 4 -x positive and not equal to zero).

We solve the equation by equivalent transformations using the rules of multiplication and division of powers:

The last transition was carried out in accordance with Theorem 1.

Answer:x= 6.

Example 3. Solve the equation:

Solution: both sides of the original equation can be divided by 0.2 x. This transition will be equivalent, since this expression is greater than zero for any value x(the exponential function is strictly positive in its domain of definition). Then the equation takes the form:

Answer: x = 0.

Example 4. Solve the equation:

Solution: we simplify the equation to an elementary one by means of equivalent transformations using the rules of division and multiplication of powers given at the beginning of the article:

Dividing both sides of the equation by 4 x, as in the previous example, is an equivalent transformation, since this expression is not equal to zero for any values x.

Answer: x = 0.

Example 5. Solve the equation:

Solution: function y = 3x, standing on the left side of the equation, is increasing. Function y = —x The -2/3 on the right side of the equation is decreasing. This means that if the graphs of these functions intersect, then at most one point. In this case, it is easy to guess that the graphs intersect at the point x= -1. There will be no other roots.

Answer: x = -1.

Example 6. Solve the equation:

Solution: we simplify the equation by means of equivalent transformations, keeping in mind everywhere that the exponential function is strictly greater than zero for any value x and using the rules for calculating the product and quotient of powers given at the beginning of the article:

Answer: x = 2.

Solving exponential inequalities

Indicative are called inequalities in which the unknown variable is contained only in exponents of some powers.

For solutions exponential inequalities knowledge of the following theorem is required:

Theorem 2. If a> 1, then the inequality a f(x) > a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x). If 0< a < 1, то показательное неравенство a f(x) > a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).

Example 7. Solve the inequality:

Solution: Let's present the original inequality in the form:

Let's divide both sides of this inequality by 3 2 x, in this case (due to the positivity of the function y= 3 2x) the inequality sign will not change:

Let's use the substitution:

Then the inequality will take the form:

So, the solution to the inequality is the interval:

moving to the reverse substitution, we get:

Due to the positivity of the exponential function, the left inequality is satisfied automatically. Using the well-known property of the logarithm, we move on to the equivalent inequality:

Since the base of the degree is a number greater than one, equivalent (by Theorem 2) is the transition to the following inequality:

So, we finally get answer:

Example 8. Solve the inequality:

Solution: Using the properties of multiplication and division of powers, we rewrite the inequality in the form:

Let's introduce a new variable:

Taking this substitution into account, the inequality takes the form:

Multiplying the numerator and denominator of the fraction by 7, we obtain the following equivalent inequality:

So, the following values ​​of the variable satisfy the inequality t:

Then, moving to the reverse substitution, we get:

Since the base of the degree here is greater than one, the transition to the inequality will be equivalent (by Theorem 2):

Finally we get answer:

Example 9. Solve the inequality:

Solution:

We divide both sides of the inequality by the expression:

It is always greater than zero (due to the positivity of the exponential function), so there is no need to change the inequality sign. We get:

t located in the interval:

Moving on to the reverse substitution, we find that the original inequality splits into two cases:

The first inequality has no solutions due to the positivity of the exponential function. Let's solve the second one:

Example 10. Solve the inequality:

Solution:

Parabola branches y = 2x+2-x 2 are directed downwards, therefore it is limited from above by the value that it reaches at its vertex:

Parabola branches y = x 2 -2x The +2 in the indicator are directed upward, which means it is limited from below by the value that it reaches at its vertex:

At the same time, the function also turns out to be bounded from below y = 3 x 2 -2x+2, which is on the right side of the equation. It reaches its smallest value at the same point as the parabola in the exponent, and this value is 3 1 = 3. So, the original inequality can only be true if the function on the left and the function on the right take on the value , equal to 3 (the intersection of the ranges of values ​​of these functions is only this number). This condition is satisfied at a single point x = 1.

Answer: x= 1.

In order to learn to decide exponential equations and inequalities, it is necessary to constantly train in solving them. Various teaching aids, problem books in elementary mathematics, collections of competitive problems, mathematics classes at school, as well as individual lessons with a professional tutor can help you in this difficult task. I sincerely wish you success in your preparation and excellent results in the exam.


Sergey Valerievich

P.S. Dear guests! Please do not write requests to solve your equations in the comments. Unfortunately, I have absolutely no time for this. Such messages will be deleted. Please read the article. Perhaps in it you will find answers to questions that did not allow you to solve your task on your own.

Exponential function is a generalization of the product of n numbers equal to a:
y (n) = a n = a·a·a···a,
to the set of real numbers x:
y (x) = ax.
Here a is a fixed real number, which is called basis of the exponential function.
An exponential function with base a is also called exponent to base a.

The generalization is carried out as follows.
For natural x = 1, 2, 3,... , the exponential function is the product of x factors:
.
Moreover, it has properties (1.5-8) (), which follow from the rules for multiplying numbers. For zero and negative values ​​of integers, the exponential function is determined using formulas (1.9-10). For fractional values ​​x = m/n rational numbers, , it is determined by formula (1.11). For real , the exponential function is defined as the limit of the sequence:
,
where is an arbitrary sequence of rational numbers converging to x: .
With this definition, the exponential function is defined for all , and satisfies properties (1.5-8), as for natural x.

A rigorous mathematical formulation of the definition of an exponential function and the proof of its properties is given on the page “Definition and proof of the properties of an exponential function”.

Properties of the Exponential Function

The exponential function y = a x has the following properties on the set of real numbers ():
(1.1) defined and continuous, for , for all ;
(1.2) for a ≠ 1 has many meanings;
(1.3) strictly increases at , strictly decreases at ,
is constant at ;
(1.4) at ;
at ;
(1.5) ;
(1.6) ;
(1.7) ;
(1.8) ;
(1.9) ;
(1.10) ;
(1.11) , .

Other useful formulas.
.
Formula for converting to an exponential function with a different exponent base:

When b = e, we obtain the expression of the exponential function through the exponential:

Private values

, , , , .

The figure shows graphs of the exponential function
y (x) = ax
for four values degree bases: a = 2 , a = 8 , a = 1/2 and a = 1/8 . It can be seen that for a > 1 the exponential function increases monotonically. The larger the base of the degree a, the stronger the growth. At 0 < a < 1 the exponential function decreases monotonically. The smaller the exponent a, the stronger the decrease.

Ascending, descending

The exponential function for is strictly monotonic and therefore has no extrema. Its main properties are presented in the table.

y = a x , a > 1 y = ax, 0 < a < 1
Domain - ∞ < x < + ∞ - ∞ < x < + ∞
Range of values 0 < y < + ∞ 0 < y < + ∞
Monotone monotonically increases monotonically decreases
Zeros, y = 0 No No
Intercept points with the ordinate axis, x = 0 y = 1 y = 1
+ ∞ 0
0 + ∞

Inverse function

The inverse of an exponential function with base a is the logarithm to base a.

If , then
.
If , then
.

Differentiation of an exponential function

To differentiate an exponential function, its base must be reduced to the number e, apply the table of derivatives and the rule for differentiating a complex function.

To do this you need to use the property of logarithms
and the formula from the derivatives table:
.

Let an exponential function be given:
.
We bring it to the base e:

Let's apply the rule of differentiation of complex functions. To do this, introduce the variable

Then

From the table of derivatives we have (replace the variable x with z):
.
Since is a constant, the derivative of z with respect to x is equal to
.
According to the rule of differentiation of a complex function:
.

Derivative of an exponential function

.
Derivative of nth order:
.
Deriving formulas > > >

An example of differentiating an exponential function

Find the derivative of a function
y = 3 5 x

Solution

Let's express the base of the exponential function through the number e.
3 = e ln 3
Then
.
Enter a variable
.
Then

From the table of derivatives we find:
.
Because the 5ln 3 is a constant, then the derivative of z with respect to x is equal to:
.
According to the rule of differentiation of a complex function, we have:
.

Answer

Integral

Expressions using complex numbers

Consider the complex number function z:
f (z) = a z
where z = x + iy; i 2 = - 1 .
Let us express the complex constant a in terms of modulus r and argument φ:
a = r e i φ
Then


.
The argument φ is not uniquely defined. In general
φ = φ 0 + 2 πn,
where n is an integer. Therefore the function f (z) is also not clear. Its main significance is often considered
.

Series expansion


.

References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.

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At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “Exponential Equations.” The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of problem, graduates can count on high scores when passing the Unified State Exam in mathematics.

Get ready for exam testing with Shkolkovo!

When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. A school textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.

The Shkolkovo educational portal invites students to use our knowledge base. We are implementing a completely new method of preparing for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.

Shkolkovo teachers collected, systematized and presented all the material necessary for successfully passing the Unified State Exam in the simplest and most accessible form.

Basic definitions and formulas are presented in the “Theoretical background” section.

To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples of exponential equations with solutions presented on this page to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or go straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.

Those examples with indicators that caused you difficulties can be added to “Favorites”. This way you can quickly find them and discuss the solution with your teacher.

To successfully pass the Unified State Exam, study on the Shkolkovo portal every day!