Question

1. log1/3 (2x) >-22. log0.5(x^2+1) меньше или равно log0.5(2x-5)

Answer 1

Task/28551058
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1.
Iog_(1/3)  (2x)  > -2   * * * ОДЗ : x >0 * * *
т.к.  основание логарифма  0 < 1/3 < 1 ,то 
0 < 2x < (1/3)⁻²    ⇔ 0< 2x < 9  ⇔ 0 < x <4,5 .

ответ: x ∈( 0 ; 4,5) .
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2.
log_(0,5)  (x²+ 1)  ≤ log_(0,5)  (2x -5)   ; ОДЗ : 2x -5 >0 ⇔  x  > 2,5 . 
основание логарифма  (0,5)  меньше единицы ,следовательно 
x²+ 1   ≥  2x -5                        > 0 ;
x² -2x +6 ≥ 0     D₁=1² -6 = -5 < 0   * * * или D=2² -4*1*6 = -20 < 0 * * *
(x -1)² +5 ≥ 0 ⇒ x ∈( -∞ ; ∞)_для всех x             * * * x ∈R * * *
учитывая ОДЗ  пишем

ответ:  x ∈( 2,5 ; ∞) .

Answer 2

To solve the inequality log₁/₃(2x) > -22 and log₀.₅(x²+1) ≤ log₀.₅(2x-5), we need to use some logarithmic properties to simplify and isolate the variable x.

1. log₁/₃(2x) > -22: Using the property that logₐ(b) > c is equivalent to a^c < b, we can rewrite the first inequality as follows: 1/3^(22) < 2x (1/3)^(22) < 2x (1/3)^22 < 2x

2. log₀.₅(x²+1) ≤ log₀.₅(2x-5): Since both sides have the same base (0.5), we can remove the logarithms and compare the expressions inside the logarithms directly: x² + 1 ≤ 2x - 5

Now we have two inequalities:

1. (1/3)^22 < 2x
2. x² + 1 ≤ 2x - 5

Let's solve each inequality separately:

1. (1/3)^22 < 2x: Calculate (1/3)^22 ≈ 1.66533454 x 10^(-13) (a very small positive value) 1.66533454 x 10^(-13) < 2x

Since 1.66533454 x 10^(-13) is essentially zero compared to 2x, we can consider it negligible. So the first inequality is effectively 0 < 2x.

2. x² + 1 ≤ 2x - 5: Move all terms to one side of the inequality: x² - 2x + 6 ≤ 0

Now, we need to find the values of x that satisfy this inequality. You can use various methods, such as factoring or the quadratic formula, to solve this quadratic inequality.

By factoring: (x - 3)(x + 2) ≤ 0

Now we have three intervals to test: x < -2, -2 ≤ x ≤ 3, and x > 3.

Testing the intervals:

For x < -2: Let's take x = -3 (a value less than -2) (-3 - 3)(-3 + 2) ≤ 0 (-6)(-1) ≤ 0 6 ≤ 0 (false)

For -2 ≤ x ≤ 3: Let's take x = 0 (a value between -2 and 3) (0 - 3)(0 + 2) ≤ 0 (-3)(2) ≤ 0 -6 ≤ 0 (true)

For x > 3: Let's take x = 4 (a value greater than 3) (4 - 3)(4 + 2) ≤ 0 (1)(6) ≤ 0 6 ≤ 0 (false)

So, the values of x that satisfy the inequality x² + 1 ≤ 2x - 5 are -2 ≤ x ≤ 3.

Now, we need to find the intersection between the solutions of both inequalities:

From the first inequality, we found that 0 < 2x. From the second inequality, we found that -2 ≤ x ≤ 3.

The common range is 0 < 2x ≤ 3.

Divide by 2 to get 0 < x ≤ 3/2.

The final solution for the inequality system is 0 < x ≤ 3/2.