Examples of systems of exponential inequalities. Exponential equations. More complex cases
On this lesson we will consider solving more complex exponential equations, recall the basic theoretical principles regarding exponential function.
1. Definition and properties of the exponential function, methods for solving the simplest exponential equations
Let us recall the definition and basic properties of the exponential function. The solution of all exponential equations and inequalities is based on these properties.
Exponential function is a function of the form , where the base is the degree and Here x is the independent variable, argument; y is the dependent variable, function.
Rice. 1. Graph of exponential function
The graph shows increasing and decreasing exponents, illustrating the exponential function with a base greater than one and less than one but greater than zero, respectively.
Both curves pass through the point (0;1)
Properties of the Exponential Function:
Domain: ;
Range of values: ;
The function is monotonic, increases with, decreases with.
A monotonic function takes each of its values given a single argument value.
When the argument increases from minus to plus infinity, the function increases from zero inclusive to plus infinity. On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero, not inclusive.
2. Solving standard exponential equations
Let us remind you how to solve the simplest exponential equations. Their solution is based on the monotonicity of the exponential function. Almost all complex exponential equations can be reduced to such equations.
The equality of exponents with equal bases is due to the property of the exponential function, namely its monotonicity.
Solution method:
Equalize the bases of degrees;
Equate the exponents.
Let's move on to consider more complex exponential equations; our goal is to reduce each of them to the simplest.
Let's get rid of the root on the left side and bring the degrees to the same base:
In order to reduce a complex exponential equation to its simplest, substitution of variables is often used.
Let's use the power property:
We are introducing a replacement. Let it be then 
Let's multiply the resulting equation by two and move all terms to the left side:
The first root does not satisfy the range of y values, so we discard it. We get:
Let's reduce the degrees to the same indicator:
Let's introduce a replacement:
Let it be then 
. With such a replacement, it is obvious that y accepts strictly positive values. We get:
We know how to solve such quadratic equations, we can write down the answer:
To make sure that the roots are found correctly, you can check using Vieta’s theorem, i.e., find the sum of the roots and their product and compare them with the corresponding coefficients of the equation.
We get:
3. Methodology for solving homogeneous exponential equations of the second degree
Let's study the following important type of exponential equations:
Equations of this type are called homogeneous of the second degree with respect to the functions f and g. On the left side there is quadratic trinomial relative to f with parameter g or quadratic trinomial relative to g with parameter f.
Solution method:
This equation can be solved as a quadratic equation, but it is easier to do it differently. There are two cases to consider:
In the first case we get
In the second case, we have the right to divide by the highest degree and get:
It is necessary to introduce a change of variables, we obtain a quadratic equation for y:
Let us note that the functions f and g can be any, but we are interested in the case when these are exponential functions.
4. Examples of solving homogeneous equations
Let's move all the terms to the left side of the equation:
Since exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when:
We get:
Let's introduce a replacement: 
(according to the properties of the exponential function)
We got a quadratic equation:
We determine the roots using Vieta’s theorem:
The first root does not satisfy the range of values of y, we discard it, we get:
Let's use the properties of degrees and reduce all degrees to simple bases:
It's easy to notice the functions f and g:
Since exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when .
and x = b is the simplest exponential equation. In him a Above zero And A does not equal one.
Solving exponential equations
From the properties of the exponential function we know that its range of values is limited to positive real numbers. Then if b = 0, the equation has no solutions. The same situation occurs in the equation where b
Now let us assume that b>0. If in the exponential function the base a is greater than unity, then the function will be increasing over the entire domain of definition. If in the exponential function for the base A the following condition is met 0 Based on this and applying the root theorem, we find that the equation a x = b has one single root, for b>0 and positive a not equal to one. To find it, you need to represent b in the form b = a c. Consider the following example: solve the equation 5 (x 2 - 2*x - 1) = 25. Let's imagine 25 as 5 2, we get: 5 (x 2 - 2*x - 1) = 5 2 . Or what is equivalent: x 2 - 2*x - 1 = 2. We solve the resulting quadratic equation using any of the known methods. We get two roots x = 3 and x = -1. Answer: 3;-1. Let's solve the equation 4 x - 5*2 x + 4 = 0. Let's make the replacement: t=2 x and get the following quadratic equation: t 2 - 5*t + 4 = 0. Now we solve the equations 2 x = 1 and 2 x = 4. Answer: 0;2. The solution to the simplest exponential inequalities is also based on the properties of increasing and decreasing functions. If in an exponential function the base a is greater than one, then the function will be increasing over the entire domain of definition. If in the exponential function for the base A the following condition is met 0 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. Exponential equations and inequalities are those in which the unknown is contained in the exponent. Solving exponential equations often comes down to solving the equation a x = a b, where a > 0, a ≠ 1, x is an unknown. This equation has a single root x = b, since the following theorem is true: Theorem. If a > 0, a ≠ 1 and a x 1 = a x 2, then x 1 = x 2. Let us substantiate the considered statement. Let us assume that the equality x 1 = x 2 does not hold, i.e. x 1< х 2 или х 1 = х 2 . Пусть, например, х 1 < х 2 . Тогда если а >1, then the exponential function y = a x increases and therefore the inequality a x 1 must be satisfied< а х 2 ; если 0 < а < 1, то функция убывает и должно выполняться неравенство а х 1 >a x 2. In both cases we received a contradiction to the condition a x 1 = a x 2. Let's consider several problems. Solve the equation 4 ∙ 2 x = 1. Solution. Let's write the equation in the form 2 2 ∙ 2 x = 2 0 – 2 x+2 = 2 0, from which we get x + 2 = 0, i.e. x = -2. Answer. x = -2. Solve equation 2 3x ∙ 3 x = 576. Solution. Since 2 3x = (2 3) x = 8 x, 576 = 24 2, the equation can be written as 8 x ∙ 3 x = 24 2 or as 24 x = 24 2. From here we get x = 2. Answer. x = 2. Solve the equation 3 x+1 – 2∙3 x - 2 = 25. Solution. Taking the common factor 3 x - 2 out of brackets on the left side, we get 3 x - 2 ∙ (3 3 – 2) = 25 – 3 x - 2 ∙ 25 = 25, whence 3 x - 2 = 1, i.e. x – 2 = 0, x = 2. Answer. x = 2. Solve the equation 3 x = 7 x. Solution. Since 7 x ≠ 0, the equation can be written as 3 x /7 x = 1, whence (3/7) x = 1, x = 0. Answer. x = 0. Solve the equation 9 x – 4 ∙ 3 x – 45 = 0. Solution. By replacing 3 x = a, this equation is reduced to the quadratic equation a 2 – 4a – 45 = 0. Solving this equation, we find its roots: a 1 = 9, and 2 = -5, whence 3 x = 9, 3 x = -5. The equation 3 x = 9 has root 2, and the equation 3 x = -5 has no roots, since the exponential function cannot take negative values. Answer. x = 2. Solving exponential inequalities often comes down to solving the inequalities a x > a b or a x< а b . Эти неравенства решаются с помощью свойства возрастания или убывания показательной функции. Let's look at some problems. Solve inequality 3 x< 81. Solution. Let's write the inequality in the form 3 x< 3 4 . Так как 3 >1, then the function y = 3 x is increasing. Therefore, for x< 4 выполняется неравенство 3 х < 3 4 , а при х ≥ 4 выполняется неравенство 3 х ≥ 3 4 . Thus, at x< 4 неравенство 3 х < 3 4 является верным, а при х ≥ 4 – неверным, т.е. неравенство Answer. X< 4.
Solve the inequality 16 x +4 x – 2 > 0. Solution. Let us denote 4 x = t, then we obtain the quadratic inequality t2 + t – 2 > 0. This inequality holds for t< -2 и при t > 1. Since t = 4 x, we get two inequalities 4 x< -2, 4 х > 1. The first inequality has no solutions, since 4 x > 0 for all x € R. We write the second inequality in the form 4 x > 4 0, whence x > 0. Answer. x > 0. Graphically solve the equation (1/3) x = x – 2/3. Solution. 1) Let’s build graphs of the functions y = (1/3) x and y = x – 2/3. 2) Based on our figure, we can conclude that the graphs of the considered functions intersect at the point with the abscissa x ≈ 1. Checking proves that x = 1 is the root of this equation: (1/3) 1 = 1/3 and 1 – 2/3 = 1/3. In other words, we have found one of the roots of the equation. 3) Let's find other roots or prove that there are none. The function (1/3) x is decreasing, and the function y = x – 2/3 is increasing. Therefore, for x > 1, the values of the first function are less than 1/3, and the second – more than 1/3; at x< 1, наоборот, значения первой функции больше 1/3, а второй – меньше 1/3. Геометрически это означает, что графики этих функций при х >1 and x< 1 «расходятся» и потому не могут иметь точек пересечения при х ≠ 1. Answer. x = 1. Note that from the solution of this problem, in particular, it follows that the inequality (1/3) x > x – 2/3 is satisfied for x< 1, а неравенство (1/3) х < х – 2/3 – при х > 1. website, when copying material in full or in part, a link to the source is required. Additional materials Teaching aids and simulators in the Integral online store for grade 11 Definition. Equations of the form: $a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ are called exponential equations. Recalling the theorems that we studied in the topic "Exponential Function", we can introduce a new theorem: B) $\sqrt(\frac(2)(3))=((\frac(2)(3)))^(\frac(1)(5))$. C) The original equation is equivalent to the equation: $x^2-6x=-3x+18$. Example. Example. Let's make a reminder of how to solve exponential equations: Example. Theorem. If $a>1$, then the exponential inequality $a^(f(x))>a^(g(x))$ is equivalent to the inequality $f(x)>g(x)$. Example. 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. C) Our inequality is equivalent to the inequality:
Then it is obvious that With will be a solution to the equation a x = a c .
We solve this equation using any of the known methods. We get the roots t1 = 1 t2 = 4Solving exponential inequalities
3 x< 81 выполняется тогда и только тогда, когда х < 4.
Lesson and presentation on the topic: "Exponential equations and exponential inequalities"
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Interactive manual for grades 9–11 "Trigonometry"
Interactive manual for grades 10–11 "Logarithms"Definition of Exponential Equations
Guys, we studied exponential functions, learned their properties and built graphs, analyzed examples of equations in which exponential functions were found. Today we will study exponential equations and inequalities.
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
Example.
Solve equations:
a) $3^(3x-3)=27$.
b) $((\frac(2)(3)))^(2x+0.2)=\sqrt(\frac(2)(3))$.
c) $5^(x^2-6x)=5^(-3x+18)$.
Solution.
a) We know well that $27=3^3$.
Let's rewrite our equation: $3^(3x-3)=3^3$.
Using the theorem above, we find that our equation reduces to the equation $3x-3=3$; solving this equation, we get $x=2$.
Answer: $x=2$.
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: $x=0$.
$x^2-3x-18=0$.
$(x-6)(x+3)=0$.
$x_1=6$ and $x_2=-3$.
Answer: $x_1=6$ and $x_2=-3$.
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: $x=0$.
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 demonstrative 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: $x=1$.
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 It is worth using if the equation, when replacing variables, simplifies its form and is much easier to solve.
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
Let's move on to inequalities. When solving inequalities, it is necessary to pay attention to the basis of the degree. There are two possible scenarios for the development of events when solving inequalities.
If $0
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:
$2x+3>4$.
$2x>1$.
$x>0.5$.
$2x-4>2$.
$x>3$.
$x^2+6x≥4x+15$.
$x^2+2x-15≥0$.
$(x-3)(x+5)≥0$.
Let's use the interval solution method:
Answer: $(-∞;-5]U)
