Question Number 172077 by Mikenice last updated on 23/Jun/22
$${solve} \\ $$$${log}_{\mathrm{4}} \left({x}+\mathrm{12}\right).{logx}^{\mathrm{2}} =\mathrm{1} \\ $$
Commented by mr W last updated on 24/Jun/22
$$\mathrm{log}\:\left({x}+\mathrm{12}\right)\mathrm{log}\:{x}^{\mathrm{2}} =\mathrm{log}\:\mathrm{4}=\mathrm{2}\:\mathrm{log}\:\mathrm{2} \\ $$$$−\mathrm{12}<{x}<\mathrm{0}: \\ $$$$\mathrm{2}\:\mathrm{log}\:\left({x}+\mathrm{12}\right)\mathrm{log}\:\left(−{x}\right)=\mathrm{2}\:\mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{log}\:\left({x}+\mathrm{12}\right)\mathrm{log}\:\left(−{x}\right)=\mathrm{log}\:\mathrm{2} \\ $$$$\mathrm{log}\:\left(−\mathrm{10}+\mathrm{12}\right)\mathrm{log}\:\left(\mathrm{10}\right)=\mathrm{log}\:\mathrm{2}\:\checkmark \\ $$$$\mathrm{log}\:\left(−\mathrm{2}+\mathrm{12}\right)\mathrm{log}\:\left(\mathrm{2}\right)=\mathrm{log}\:\mathrm{2}\:\checkmark \\ $$$${x}=−\mathrm{10},\:−\mathrm{2}\:{are}\:{roots}. \\ $$$${x}>\mathrm{0}: \\ $$$$\mathrm{log}\:\left({x}+\mathrm{12}\right)\mathrm{log}\:\left({x}\right)=\mathrm{log}\:\mathrm{2} \\ $$$${we}\:{can}\:{only}\:{approximate}.\:{x}\approx\mathrm{1}.\mathrm{8358} \\ $$