Question Number 163060 by tounghoungko last updated on 03/Jan/22
$$\:\:\mathrm{2log}\:_{\mathrm{3}} \left(\frac{{x}^{\mathrm{2}} }{\mathrm{27}}\right)\:=\:\mathrm{2}+\:\frac{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{log}\:_{\mathrm{5}} \left(\sqrt{{x}}\:\right)}\: \\ $$
Commented by tounghoungko last updated on 04/Jan/22
Answered by aleks041103 last updated on 04/Jan/22
$${log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−{log}_{\mathrm{3}} \left(\mathrm{27}\right)=\mathrm{1}−\frac{{log}_{\mathrm{3}} {x}}{\mathrm{2}{log}_{\mathrm{5}} \sqrt{{x}}} \\ $$$${log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−\mathrm{3}=\mathrm{1}−\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{5}} {x}} \\ $$$${log}_{\mathrm{5}} {x}=\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{5}} \\ $$$$\Rightarrow{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−\mathrm{4}=−\frac{{log}_{\mathrm{3}} {x}}{\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{5}}}=−{log}_{\mathrm{3}} \mathrm{5} \\ $$$$\Rightarrow{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)=\mathrm{4}−{log}_{\mathrm{3}} \mathrm{5}= \\ $$$$={log}_{\mathrm{3}} \mathrm{3}^{\mathrm{4}} −{log}_{\mathrm{3}} \mathrm{5}={log}_{\mathrm{3}} \left(\frac{\mathrm{81}}{\mathrm{5}}\right) \\ $$$$\Rightarrow{x}=\sqrt{\frac{\mathrm{81}}{\mathrm{5}}}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{5}}} \\ $$$${Ans}.\:{x}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{5}}} \\ $$
Commented by mkam last updated on 03/Jan/22
$${how}\:{log}_{\mathrm{5}} {x}\:=\:\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{5}}\:? \\ $$
Commented by aleks041103 last updated on 03/Jan/22
$${b}={a}^{{log}_{{a}} {b}} =\left({c}^{{log}_{{c}} {a}} \right)^{{log}_{{a}} {b}} ={c}^{{log}_{{c}} {b}} \\ $$$$\Rightarrow{log}_{{c}} {b}={log}_{{c}} {a}\:×\:{log}_{{a}} {b} \\ $$$$\Rightarrow{log}_{{a}} {b}=\frac{{log}_{{c}} {b}}{{log}_{{c}} {a}} \\ $$