Question

Помогите пж со вторым заданием

Answer 1

$1) \ \text{log}_{2}x = -3\\x = 2^{-3} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$

$\text{OTBET:} \ x = \dfrac{1}{8}$

$2) \ \text{log}_{6}(x^{2} + 3x - 8) = \text{log}_{6}2\\x^{2} + 3x - 8 = 2\\x^{2} + 3x - 10 = 0\\x_{1} = -5; \ x_{2} = 2$

$\text{OTBET:} \ x_{1} = -5; \ x_{2} = 2$