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3-log-3-2-log-3-3log-1-3-cot-pi-3-log-pi-3-log-2-pi-




Question Number 95968 by i jagooll last updated on 29/May/20
3^(((log _3 (2)+log _3 (3log _(1/3) (cot (π/3))))/(log _π (3).(log _2 (π)))) ? )
$$\mathrm{3}^{\frac{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3log}\:_{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{cot}\:\frac{\pi}{\mathrm{3}}\right)\right)}{\mathrm{log}\:_{\pi} \left(\mathrm{3}\right).\left(\mathrm{log}\:_{\mathrm{2}} \left(\pi\right)\right)}\:?\:} \\ $$
Answered by john santu last updated on 29/May/20
⇒log _3 (2)+log _3 (3log _3 (tan ((π/3)))=  log _3 (2)+log _3 (3log _3 ((√3)))=  log _3 (2×(3/2)) = 1  ⇒(1/(log _π (3).log _2 (π))) = (1/(log _2 (3)))  so 3^(log _3 (2))  = 2 .
$$\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3log}\:_{\mathrm{3}} \left(\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}\right)\right)=\right. \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3log}\:_{\mathrm{3}} \left(\sqrt{\mathrm{3}}\right)\right)= \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{\pi} \left(\mathrm{3}\right).\mathrm{log}\:_{\mathrm{2}} \left(\pi\right)}\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}\right)} \\ $$$$\mathrm{so}\:\mathrm{3}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)} \:=\:\mathrm{2}\:. \\ $$
Commented by 1549442205 last updated on 29/May/20
your answer is false,numerator equal   to 1!
$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{false},\mathrm{numerator}\:\mathrm{equal}\: \\ $$$$\mathrm{to}\:\mathrm{1}! \\ $$
Commented by 1549442205 last updated on 29/May/20
log_3 (2)+log_3 (3log_(1/3) (cot(π/3)))=  log_3 (2)+log_3 (3log_(1/3) ((1/3))^(1/2) )=log_3 (2)+  log_3 (3.(1/2))=log_3 (2)+log_3 3−log_3 (2)=1
$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{3}} \left(\mathrm{3log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{cot}\frac{\pi}{\mathrm{3}}\right)\right)= \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{3}} \left(\mathrm{3log}_{\frac{\mathrm{1}}{\mathrm{3}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)=\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+ \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{3}.\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{lo}\underset{\mathrm{3}} {\mathrm{g}3}−\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by i jagooll last updated on 29/May/20
yes...thank you
$$\mathrm{yes}…\mathrm{thank}\:\mathrm{you} \\ $$

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