Question

Помогите пж со вторым заданием

Answer 1

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

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

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

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