Question Number 90806 by john santu last updated on 26/Apr/20
$$\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:<\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mid{x}−\mathrm{2}\mid} \left(\mathrm{2}−{x}\right)^{\mathrm{2}} \\ $$
Commented by john santu last updated on 26/Apr/20
$$\Rightarrow\left(\mathrm{2}−{x}\right)^{\mathrm{2}} =\:\mid\mathrm{2}−{x}\mid^{\mathrm{2}} \:=\:\mid{x}−\mathrm{2}\mid^{\mathrm{2}} \\ $$$$\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:<\:\mathrm{log}_{\mid{x}−\mathrm{2}\mid} \mid{x}−\mathrm{2}\mid \\ $$$$\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:<\:\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\:−\mathrm{log}_{\left({x}+\mathrm{4}\right)} \left({x}+\mathrm{4}\right)<\:\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}−{x}−\mathrm{4}}{{x}+\mathrm{4}−\mathrm{1}}\:<\:\mathrm{0} \\ $$$$\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{8}\right)}{\left({x}+\mathrm{3}\right)}\:<\:\mathrm{0}\: \\ $$$$\left({ii}\right)\:{x}\neq\mathrm{2}\:,\:{x}\neq\mathrm{3}\:,\:{x}\neq\mathrm{1}\: \\ $$$${solution}\:{x}\in\:\left(−\mathrm{4};−\mathrm{3}\right)\:\cup\:\left(\mathrm{1};\mathrm{2}\right)\:\cup\:\left(\mathrm{6};\mathrm{8}\right)\: \\ $$