Question Number 95968 by i jagooll last updated on 29/May/20
$$\mathrm{3}^{\frac{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3log}\:_{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{cot}\:\frac{\pi}{\mathrm{3}}\right)\right)}{\mathrm{log}\:_{\pi} \left(\mathrm{3}\right).\left(\mathrm{log}\:_{\mathrm{2}} \left(\pi\right)\right)}\:?\:} \\ $$
Answered by john santu last updated on 29/May/20
$$\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3log}\:_{\mathrm{3}} \left(\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}\right)\right)=\right. \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3log}\:_{\mathrm{3}} \left(\sqrt{\mathrm{3}}\right)\right)= \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{\pi} \left(\mathrm{3}\right).\mathrm{log}\:_{\mathrm{2}} \left(\pi\right)}\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)} \\ $$$$\mathrm{so}\:\mathrm{3}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)} \:=\:\mathrm{2}\:. \\ $$
Commented by 1549442205 last updated on 29/May/20
$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{false},\mathrm{numerator}\:\mathrm{equal}\: \\ $$$$\mathrm{to}\:\mathrm{1}! \\ $$
Commented by 1549442205 last updated on 29/May/20
$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{3}} \left(\mathrm{3log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{cot}\frac{\pi}{\mathrm{3}}\right)\right)= \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{3}} \left(\mathrm{3log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)=\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+ \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{3}.\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{lo}\underset{\mathrm{3}} {\mathrm{g}3}−\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by i jagooll last updated on 29/May/20
$$\mathrm{yes}…\mathrm{thank}\:\mathrm{you} \\ $$