Question Number 163061 by tounghoungko last updated on 03/Jan/22
$$\:\sqrt[{\mathrm{log}\:_{{x}} \left(\frac{\mathrm{243}}{{x}}\right)}]{\mathrm{2}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{log}\:_{{x}} \left(\frac{{x}^{\mathrm{5}} }{\mathrm{9}}\right)}\: \\ $$
Answered by mahdipoor last updated on 03/Jan/22
$${get}\:\mathrm{5}−\mathrm{2}{log}_{{x}} \mathrm{3}={u} \\ $$$$\begin{cases}{{log}_{{x}} \left(\frac{\mathrm{243}}{{x}}\right)=\mathrm{5}{log}_{{x}} \mathrm{3}−\mathrm{1}=\frac{−\mathrm{5}}{\mathrm{2}}{u}+\frac{\mathrm{23}}{\mathrm{2}}\:}\\{{log}_{{x}} \left(\frac{{x}^{\mathrm{5}} }{\mathrm{9}}\right)=\mathrm{5}−\mathrm{2}{log}_{{x}} \mathrm{3}={u}}\end{cases} \\ $$$$\Rightarrow\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{11}.\mathrm{5}−\mathrm{2}.\mathrm{5}{u}\right)} ={u}^{\frac{\mathrm{1}}{\mathrm{3}}} \Rightarrow\:{get}\:\mathrm{2}^{\mathrm{1}.\mathrm{5}} ={e}^{\mathrm{0}.\mathrm{4}{A}} \\ $$$$\Rightarrow{e}^{{A}\left(\mathrm{4}.\mathrm{6}−{u}\right)} ={u}\Rightarrow{e}^{\mathrm{4}.\mathrm{6}{A}} ={ue}^{{Au}} \Rightarrow{Ae}^{\mathrm{4}.\mathrm{6}{A}} ={Aue}^{{Au}} \\ $$$${Au}={w}\left({Ae}^{\mathrm{4}.\mathrm{6}{A}} \right)\Rightarrow\mathrm{5}−\mathrm{2}{log}_{{x}} \mathrm{3}=\frac{\mathrm{1}}{{A}}{w}\left({Ae}^{\mathrm{4}.\mathrm{6}{A}} \right) \\ $$$$\Rightarrow{x}=\mathrm{9}^{\left(\frac{{A}}{\mathrm{5}{A}−{W}}\right)} \:\:\begin{cases}{{A}=\frac{\mathrm{15}}{\mathrm{4}}{ln}\mathrm{2}}\\{{W}={w}\left({Ae}^{\mathrm{4}.\mathrm{6}{A}} \right)}\end{cases} \\ $$