How to solve exponential equations and inequalities. Solving exponential equations and inequalities

Consider an example: solve inequality (0.5) (7 - 3*x)< 4.

Note that 4 = (0.5) 2 . Then the inequality will take the form (0.5)(7 - 3*x)< (0.5) (-2) . Основание показательной функции 0.5 меньше единицы, следовательно, она убывает. В этом случае надо поменять знак неравенства и не записывать только показатели.

We get: 7 - 3*x>-2.

Hence: x<3.

Answer: x<3.

If the base in the inequality was greater than one, then when getting rid of the base, there would be no need to change the sign of the inequality.

Lesson and presentation on the topic: "Exponential equations and exponential inequalities"

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Definition of Exponential Equations

Definition. Equations of the form: $a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ are called exponential equations.

Remembering the theorems that we studied in the topic " Exponential function", we can introduce a new theorem:
Theorem. The exponential equation $a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ is equivalent to the equation $f(x)=g(x)$.

Examples of exponential equations

B) $\sqrt(\frac(2)(3))=((\frac(2)(3)))^(\frac(1)(5))$.
Then our equation can be rewritten: $((\frac(2)(3)))^(2x+0.2)=((\frac(2)(3)))^(\frac(1)(5) )=((\frac(2)(3)))^(0.2)$.
$2х+0.2=0.2$.
$x=0$.
Answer: $x=0$.

C) The original equation is equivalent to the equation: $x^2-6x=-3x+18$.
$x^2-3x-18=0$.
$(x-6)(x+3)=0$.
$x_1=6$ and $x_2=-3$.
Answer: $x_1=6$ and $x_2=-3$.

Example.
Solve the equation: $\frac(((0.25))^(x-0.5))(\sqrt(4))=16*((0.0625))^(x+1)$.
Solution:
Let's perform a series of actions sequentially and bring both sides of our equation to the same bases.
Let's perform a number of operations on the left side:
1) $((0.25))^(x-0.5)=((\frac(1)(4)))^(x-0.5)$.
2) $\sqrt(4)=4^(\frac(1)(2))$.
3) $\frac(((0.25))^(x-0.5))(\sqrt(4))=\frac(((\frac(1)(4)))^(x-0 ,5))(4^(\frac(1)(2)))= \frac(1)(4^(x-0.5+0.5))=\frac(1)(4^x) =((\frac(1)(4)))^x$.
Let's move on to the right side:
4) $16=4^2$.
5) $((0.0625))^(x+1)=\frac(1)((16)^(x+1))=\frac(1)(4^(2x+2))$.
6) $16*((0.0625))^(x+1)=\frac(4^2)(4^(2x+2))=4^(2-2x-2)=4^(-2x )=\frac(1)(4^(2x))=((\frac(1)(4)))^(2x)$.
The original equation is equivalent to the equation:
$((\frac(1)(4)))^x=((\frac(1)(4)))^(2x)$.
$x=2x$.
$x=0$.
Answer: $x=0$.

Example.
Solve the equation: $9^x+3^(x+2)-36=0$.
Solution:
Let's rewrite our equation: $((3^2))^x+9*3^x-36=0$.
$((3^x))^2+9*3^x-36=0$.
Let's make a change of variables, let $a=3^x$.
In the new variables, the equation will take the form: $a^2+9a-36=0$.
$(a+12)(a-3)=0$.
$a_1=-12$ and $a_2=3$.
Let's perform the reverse change of variables: $3^x=-12$ and $3^x=3$.
In the last lesson we learned that exponential expressions can only take positive values, remember the graph. This means that the first equation has no solutions, the second equation has one solution: $x=1$.
Answer: $x=1$.

Let's make a reminder of how to solve exponential equations:
1. Graphic method. We represent both sides of the equation in the form of functions and build their graphs, find the points of intersection of the graphs. (We used this method in the last lesson).
2. The principle of equality of indicators. The principle is based on the fact that two expressions with the same bases are equal if and only if the degrees (exponents) of these bases are equal. $a^(f(x))=a^(g(x))$ $f(x)=g(x)$.
3. Variable replacement method. This method should be used if the equation, when replacing variables, simplifies its form and is much easier to solve.

Example.
Solve the system of equations: $\begin (cases) (27)^y*3^x=1, \\ 4^(x+y)-2^(x+y)=12. \end (cases)$.
Solution.
Let's consider both equations of the system separately:
$27^y*3^x=1$.
$3^(3y)*3^x=3^0$.
$3^(3y+x)=3^0$.
$x+3y=0$.
Consider the second equation:
$4^(x+y)-2^(x+y)=12$.
$2^(2(x+y))-2^(x+y)=12$.
Let's use the change of variables method, let $y=2^(x+y)$.
Then the equation will take the form:
$y^2-y-12=0$.
$(y-4)(y+3)=0$.
$y_1=4$ and $y_2=-3$.
Let's move on to the initial variables, from the first equation we get $x+y=2$. The second equation has no solutions. Then our initial system of equations is equivalent to the system: $\begin (cases) x+3y=0, \\ x+y=2. \end (cases)$.
Subtract the second from the first equation, we get: $\begin (cases) 2y=-2, \\ x+y=2. \end (cases)$.
$\begin (cases) y=-1, \\ x=3. \end (cases)$.
Answer: $(3;-1)$.

Exponential inequalities

Theorem. If $a>1$, then the exponential inequality $a^(f(x))>a^(g(x))$ is equivalent to the inequality $f(x)>g(x)$.
If $0 a^(g(x))$ is equivalent to the inequality $f(x)

Example.
Solve inequalities:
a) $3^(2x+3)>81$.
b) $((\frac(1)(4)))^(2x-4) c) $(0.3)^(x^2+6x)≤(0.3)^(4x+15)$ .
Solution.
a) $3^(2x+3)>81$.
$3^(2x+3)>3^4$.
Our inequality is equivalent to inequality:
$2x+3>4$.
$2x>1$.
$x>0.5$.

B) $((\frac(1)(4)))^(2x-4) $((\frac(1)(4)))^(2x-4) In our equation, the base is when the degree is less than 1, then When replacing an inequality with an equivalent one, it is necessary to change the sign.
$2x-4>2$.
$x>3$.

C) Our inequality is equivalent to the inequality:
$x^2+6x≥4x+15$.
$x^2+2x-15≥0$.
$(x-3)(x+5)≥0$.
Let's use the interval solution method:
Answer: $(-∞;-5]U)