Inequalities with different degrees. Exponential inequalities

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 it out of brackets on the left side common multiplier 3 x - 2, 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 given equation reduces 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.

Solution 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 exponential function.

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
3 x< 81 выполняется тогда и только тогда, когда х < 4.

Answer. X< 4.

Solve the inequality 16 x +4 x – 2 > 0.

Solution.

Let us denote 4 x = t, then we get 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.

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and x = b is the simplest exponential equation. In it a greater than 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 completed next condition 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.
Then it is obvious that With will be a solution to the equation a x = a c .

Let's consider next example: solve 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.

Solving what we got quadratic equation any of 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.
We solve this equation using any of the known methods. We get the roots t1 = 1 t2 = 4

Now we solve the equations 2 x = 1 and 2 x = 4.

Answer: 0;2.

Solving exponential inequalities

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, then this function will be decreasing on the entire set of real numbers.