Question

Log0.7(x У 2 степени -4x-5)=log0.7(7-3x)

Answer 1

log₀,₇(x²-4x-5)=log₀,₇(7-3x)

ОДЗ:  x²-4x-5>0    x²-5x+x-5>0       x*(x-5)+(x-5)>0    (x-5)(x+1)>0

-∞____+____-1____-____5____+____+∞

x∈(-∞;-1)U(5;+∞)

7-3x>0        3x<7        x<7/3       x<2¹/₃       ⇒          x∈(-∞;-1).  

x²-4x-5=7-3x

x²-x-12=0    D=49        √D=7

x₁=4  ∉ОДЗ    

x₂=-3 ∈ОДЗ

Ответ: x=-3.

Answer 2

To solve the equation log₀.₇(x² - 4x - 5) = log₀.₇(7 - 3x), we can set the logarithmic expressions equal to each other:

x² - 4x - 5 = 7 - 3x

Next, we can bring all the terms to one side to form a quadratic equation:

x² - 4x + 3x - 7 + 5 = 0

Simplifying the equation:

x² - x - 2 = 0

Now we can factor the quadratic equation:

(x - 2)(x + 1) = 0

Setting each factor equal to zero:

x - 2 = 0 or x + 1 = 0

Solving for x:

x = 2 or x = -1

Therefore, the solutions to the equation log₀.₇(x² - 4x - 5) = log₀.₇(7 - 3x) are x = 2 and x = -1.