Question Number 161362 by cortano last updated on 17/Dec/21
$$\:\mathrm{log}\:_{\sqrt{\frac{{x}}{\mathrm{3}}}} \left(\mathrm{3}{x}−\mathrm{54}\right)^{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)} \:=\:\mathrm{18}−\mathrm{3log}\:_{\frac{{x}}{\mathrm{3}}} \left({x}^{\mathrm{2}} \right) \\ $$$$\:{x}=? \\ $$
Commented by bobhans last updated on 17/Dec/21
![((log _3 (x)[1+log _3 (x−18)])/((1/2)[log _3 (x)−1])) = 18−((6 log _3 (x))/(log _3 (x)−1)) ((2 log _3 (x)[1+log _3 (x−18)])/(log _3 (x)−1)) = ((12 log _3 (x)−18)/(log _3 (x)−1)) 2log _3 (x)+2log _3 (x)log _3 (x−18)=12log _3 (x)−18 log _3 (x)log _3 (x−18)−5log _3 (x)+9 = 0](https://www.tinkutara.com/question/Q161400.png)
$$\:\frac{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\left[\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}−\mathrm{18}\right)\right]}{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{1}\right]}\:=\:\mathrm{18}−\frac{\mathrm{6}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{1}} \\ $$$$\:\frac{\mathrm{2}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\left[\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}−\mathrm{18}\right)\right]}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{1}}\:=\:\frac{\mathrm{12}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{18}}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{1}} \\ $$$$\:\mathrm{2log}\:_{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{2log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}−\mathrm{18}\right)=\mathrm{12log}\:_{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{18} \\ $$$$\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}−\mathrm{18}\right)−\mathrm{5log}\:_{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{9}\:=\:\mathrm{0} \\ $$$$\: \\ $$