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Question Number 90806 by john santu last updated on 26/Apr/20

log_((x+4)) (x^2 −8x+12) < (1/2)log_(∣x−2∣) (2−x)^2

$$\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:<\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mid{x}−\mathrm{2}\mid} \left(\mathrm{2}−{x}\right)^{\mathrm{2}} \\ $$

Commented byjohn santu last updated on 26/Apr/20

⇒(2−x)^2 = ∣2−x∣^2 = ∣x−2∣^2 log_((x+4)) (x^2 −8x+12) < log_(∣x−2∣) ∣x−2∣ log_((x+4)) (x^2 −8x+12) < 1 ⇒ log_((x+4)) (x^2 −8x+12) −log_((x+4)) (x+4)< 0 ((x^2 −8x+12−x−4)/(x+4−1)) < 0 (((x−1)(x−8))/((x+3))) < 0 (ii) x≠2 , x≠3 , x≠1 solution x∈ (−4;−3) ∪ (1;2) ∪ (6;8)

$$\Rightarrow\left(\mathrm{2}−{x}\right)^{\mathrm{2}} =\:\mid\mathrm{2}−{x}\mid^{\mathrm{2}} \:=\:\mid{x}−\mathrm{2}\mid^{\mathrm{2}} \\ $$ $$\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:<\:\mathrm{log}_{\mid{x}−\mathrm{2}\mid} \mid{x}−\mathrm{2}\mid \\ $$ $$\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:<\:\mathrm{1}\: \\ $$ $$\Rightarrow\:\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:−\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}+\mathrm{4}\right)<\:\mathrm{0} \\ $$ $$\frac{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}−{x}−\mathrm{4}}{{x}+\mathrm{4}−\mathrm{1}}\:<\:\mathrm{0} \\ $$ $$\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{8}\right)}{\left({x}+\mathrm{3}\right)}\:<\:\mathrm{0}\: \\ $$ $$\left({ii}\right)\:{x}\neq\mathrm{2}\:,\:{x}\neq\mathrm{3}\:,\:{x}\neq\mathrm{1}\: \\ $$ $${solution}\:{x}\in\:\left(−\mathrm{4};−\mathrm{3}\right)\:\cup\:\left(\mathrm{1};\mathrm{2}\right)\:\cup\:\left(\mathrm{6};\mathrm{8}\right)\: \\ $$

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