Question

Log0,7(2x-8)>log0,7(x^2-4)

Answer 1

Log₀,₇(2x-8)>log₀,₇(x²-4)    
ОДЗ: 2x-8>0   x>4   x²-4>0  (x-2)(x+2)>0 x∈(-∞;-2)U(2;+∞) ⇒ x∈(4;+∞)
2x-8<x²-4
x²-2x+4>0
(x²-2x+1)+3>0
(x-1)²+3>0  ⇒
x∈(-∞;+∞).
Учитывая ОДЗ: x∈(4;+∞).

Answer 2

Log₀,₇(2x-8)>log₀,₇(x²-4)  
 
2x-8>0                  x²-4>0 
x>4                       (x-2)(x+2)>0
х∈(4;+∞).              x∈(-∞;-2)U(2;+∞)
 
Т.к. 0,7<1, то функция убывающая и
2x-8<x²-4
x²-2x+4>0
(x²-2x+1)+3>0
(x-1)²+3>0  при любом х.
x∈(-∞;+∞).

С учетом ограничений: x∈(4;+∞).

Answer 3

To solve this inequality, we need to use the properties of logarithms. In particular, we can use the fact that for any base a and positive numbers x and y:

loga(xy) = loga(x) + loga(y) loga(x/y) = loga(x) - loga(y) loga(x^n) = n*loga(x)

Using these properties, we can rewrite the inequality as follows:

log0,7(2x-8) > log0,7(x^2-4) log0,7[(2x-8)/(x^2-4)] > 0 (using the second property above)

Now, we need to find the values of x that make the expression inside the logarithm greater than 0. We can factor the numerator and denominator:

(2x-8)/(x^2-4) = 2(x-4)/[(x-2)(x+2)]

The expression is positive when the numerator and denominator have the same sign, or when they are both zero. Therefore, we need to solve the following inequalities:

2(x-4) > 0 (when the denominator is positive) (x-2)(x+2) < 0 (when the denominator is negative)

Solving the first inequality, we get:

x > 2

Solving the second inequality, we can use a sign chart or test values to find the solution set:

(x-2)(x+2) < 0 (x-2) < 0 and (x+2) > 0 or (x-2) > 0 and (x+2) < 0 x < -2 or 2 < x < 4

Therefore, the solution set is:

x < -2 or 2 < x < 4

However, we need to check whether the solution set satisfies the original inequality. Since the logarithm function is only defined for positive arguments, we need to ensure that the expression inside the logarithms is positive. We can do this by checking the endpoints and critical points of the solution set:

x = -3: the expression inside the logarithm is negative, so this value is not a solution. x = -2: the expression inside the logarithm is undefined, so we exclude this value from the solution set. x = 2: the expression inside the logarithm is undefined, so we exclude this value from the solution set. x = 4: the expression inside the logarithm is positive, so this value is a solution.

Therefore, the final solution set is:

x ∈ (2, 4]