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

∣1−log _(((1/6))) (x)∣ +2 = ∣3 −log _(((1/6))) (3)∣

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

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

let 1+log _6 (x) = t ∣t∣ + 2 = ∣2+t ∣ t^2 + 4∣t∣ +4 = 4+4t+t^2 ∣t∣ − t = 0 case(1) t ≥ 0 ⇒ t−t = 0 , 0 = 0 always true , so t ≥ 0 is solution 1+log _6 (x) ≥ 0 ⇒ log _6 (x) ≥−1 x ≥ (1/6) ⇒ solution x ∈ [(1/6); +∞)

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

Commented by bobhans last updated on 23/May/20

thank you

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

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