Question Number 221760 by OmoloyeMichael last updated on 09/Jun/25
Answered by shunmisaki007 last updated on 09/Jun/25
$$\left(\mathrm{For}\:{x}>\mathrm{0}\:\mathrm{and}\:{x}\neq\mathrm{1}.\right) \\ $$$$\mathrm{log}_{{x}} \left(\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}\right)^{\mathrm{log}_{\mathrm{3}} \left({x}\right)} =\mathrm{2} \\ $$$$\mathrm{log}_{\mathrm{3}} \left({x}\right)\mathrm{log}_{{x}} \left(\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}\right)=\mathrm{2} \\ $$$$\frac{\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left(\mathrm{3}\right)}\:\frac{\mathrm{ln}\left(\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}\right)}{\mathrm{ln}\left({x}\right)}=\mathrm{2} \\ $$$$\mathrm{ln}\left(\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}\right)=\mathrm{2ln}\left(\mathrm{3}\right)=\mathrm{ln}\left(\mathrm{3}^{\mathrm{2}} \right)=\mathrm{ln}\left(\mathrm{9}\right) \\ $$$$\frac{\mathrm{log}_{\mathrm{4}} \left({x}\right)}{\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}}=\mathrm{9} \\ $$$$\mathrm{log}_{\mathrm{4}} \left({x}\right)=\mathrm{9log}_{\mathrm{4}} \left({x}\right)−\mathrm{27} \\ $$$$\left(\mathrm{Condition}:\:\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{3}\neq\mathrm{0}\:\right. \\ $$$$\:\:\:\Rightarrow\mathrm{log}_{\mathrm{4}} \left({x}\right)\neq\mathrm{3}\Rightarrow\mathrm{ln}\left({x}\right)\neq\mathrm{3ln}\left(\mathrm{4}\right) \\ $$$$\left.\:\:\:\Rightarrow{x}\neq\mathrm{64}.\right) \\ $$$$\mathrm{8log}_{\mathrm{4}} \left({x}\right)=\mathrm{27} \\ $$$$\mathrm{ln}\left({x}\right)=\frac{\mathrm{27}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{4}\right) \\ $$$${x}=\mathrm{4}^{\frac{\mathrm{27}}{\mathrm{8}}} =\mathrm{2}^{\frac{\mathrm{27}}{\mathrm{4}}} \:\boldsymbol{{Ans}}. \\ $$