Область определения:
x > 0
x ∈ (0; +oo)
[m]\frac{1}{\log_2 x} + 8\log_2 x + 9 = 0[/m]
Замена t = log_2 x
[m]\frac{1}{t} + 8t + 9 = 0[/m]
8t^2 + 9t + 1 = 0
(t + 1)(8t + 1) = 0
[m]t_1 = \log_2 x = -1[/m]
[m]x_1 = 2^{-1} = \frac{1}{2}[/m]
[m]t_2 = \log_2 x = -\frac{1}{8}[/m]
[m]x_2 = 2^{-1/8} = \frac{1}{\sqrt[8]{2}}[/m]