Question-174884

Question Number 174884 by daus last updated on 13/Aug/22

Answered by aleks041103 last updated on 13/Aug/22

({l o g}_{b} a) ({l o g}_{a} b) = {l o g}_{a} (b^{{l o g}_{b} a}) = {l o g}_{a} a = 1

\Rightarrow {l o g}_{x} 2 = \frac{1}{{l o g}_{2} x}

l e t {l o g}_{2} x = t

1 + t - \frac{6}{t} > 0

\frac{t^{2} + t - 6}{t} > 0

\frac{(t + 3) (t - 2)}{t} > 0

t_{1, 2, 3} = - 3, 0, 2

\Rightarrow t = {l o g}_{2} x \in (- 3, 0) \cup (2, \infty)

\Rightarrow x \in (\frac{1}{8}, 1) \cup (4, \infty)