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Question Number 95108 by bobhans last updated on 23/May/20

$$\mid 1-\log {}_{\left(\frac{1}{6}\right)}\left(\mathrm{x}\right)\mid +2=\mid 3-\log {}_{\left(\frac{1}{6}\right)}\left(3\right)\mid$$

Answered by john santu last updated on 23/May/20

$$\mathrm{let}1+\log {}_6\left(\mathrm{x}\right)=\mathrm{t}$$$$\mid \mathrm{t}\mid +2=\mid 2+\mathrm{t}\mid$$$${\mathrm{t}}^2+4\mid \mathrm{t}\mid +4=4+4\mathrm{t}+{\mathrm{t}}^2$$$$\mid \mathrm{t}\mid -\mathrm{t}=0$$$$\mathrm{case}\left(1\right)\mathrm{t}⩾0\Rightarrow \mathrm{t}-\mathrm{t}=0,0=0$$$$\mathrm{always}\mathrm{true},\mathrm{so}\mathrm{t}⩾0\mathrm{is}\mathrm{solution}$$$$1+\log {}_6\left(x\right)⩾0\Rightarrow \log {}_6\left(x\right)⩾-1$$$$x⩾\frac{1}{6}\Rightarrow \mathrm{solution}x\in [\frac{1}{6};+\mathrm{\infty })$$

Commented by bobhans last updated on 23/May/20

$$\mathrm{thank}\mathrm{you}$$

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