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

Answered by aleks041103 last updated on 13/Aug/22

$$\left({log}_{b}a\right)\left({log}_{a}b\right)={log}_a\left(b^{{log}_{b}a}\right)={log}_{a}a=1$$$$\Rightarrow {log}_{x}2=\frac{1}{{log}_{2}x}$$$$let{log}_{2}x=t$$$$1+t-\frac{6}{t}>0$$$$\frac{t^2+t-6}{t}>0$$$$\frac{(t+3)(t-2)}{t}>0$$$$t_{1,2,3}=-3,0,2$$$$\Rightarrow t={log}_{2}x\in (-3,0)\cup (2,\mathrm{\infty })$$$$\Rightarrow x\in (\frac{1}{8},1)\cup (4,\mathrm{\infty })$$

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