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…
Question Number 97412 by mr W last updated on 07/Jun/20 $${prove} \\ $$$$\sqrt{\mathrm{2}}<\mathrm{log}_{\mathrm{2}} \:\mathrm{3}<\sqrt{\mathrm{3}} \\ $$ Answered by bobhans last updated on 08/Jun/20 $$\sqrt{\mathrm{8}}\:<\:\sqrt{\mathrm{9}}\:\Rightarrow\mathrm{log}\:_{\mathrm{2}} \left(\sqrt{\mathrm{8}}\right)\:<\:\mathrm{log}\:_{\mathrm{2}}…
Question Number 97390 by shaxzod last updated on 07/Jun/20 $${compare}\:{log}_{\mathrm{2}} \mathrm{3}\:{with}\:{log}_{\mathrm{3}} \mathrm{4} \\ $$ Answered by mr W last updated on 07/Jun/20 $$\mathrm{9}>\mathrm{8} \\ $$$$\mathrm{3}^{\mathrm{2}}…
Question Number 97173 by 675480065 last updated on 06/Jun/20 Answered by Sourav mridha last updated on 06/Jun/20 $$\:\boldsymbol{{I}}=\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{+\pi}{\mathrm{4}}} \frac{\boldsymbol{{dx}}}{\left(\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{sinx}}} \right)\left(\mathrm{2}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right)}…\left(\boldsymbol{{i}}\right) \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{also}}\:\boldsymbol{{I}}=\frac{\mathrm{2}}{\boldsymbol{\pi}}\int_{+\frac{\pi}{\mathrm{4}}} ^{−\frac{\boldsymbol{\pi}}{\mathrm{4}}} \frac{\mathrm{d}\left(−\mathrm{x}\right)}{\left(\mathrm{1}+\boldsymbol{{e}}^{\boldsymbol{{sin}}\left(−\boldsymbol{{x}}\right)}…
Question Number 96636 by O Predador last updated on 03/Jun/20 Commented by hknkrc46 last updated on 03/Jun/20 $$\mathrm{5}^{{x}} −\left(\mathrm{9},\mathrm{8}\right)^{{x}} =\mathrm{7}^{{x}} \:\Rightarrow\:\mathrm{5}^{{x}} −\left(\frac{\mathrm{49}}{\mathrm{5}}\right)^{{x}} =\mathrm{7}^{{x}} \\ $$$$\Rightarrow\:\mathrm{5}^{{x}}…
Question Number 96508 by bemath last updated on 02/Jun/20 Answered by 1549442205 last updated on 02/Jun/20 $$\boldsymbol{\mathrm{F}}\mathrm{rom}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have}\:\mathrm{b}=\mathrm{a}^{\frac{\mathrm{3}}{\mathrm{2}}} =\sqrt{\mathrm{a}^{\mathrm{3}} }=\mathrm{a}\sqrt{\mathrm{a}} \\ $$$$\mathrm{d}=\mathrm{c}^{\frac{\mathrm{5}}{\mathrm{4}}} =^{\mathrm{4}} \sqrt{\mathrm{c}^{\mathrm{5}} }=\mathrm{c}^{\mathrm{4}} \sqrt{\mathrm{c}}\Rightarrow\mathrm{b}−\mathrm{d}=\mathrm{a}\sqrt{\mathrm{a}}−\mathrm{c}^{\mathrm{4}}…
Question Number 161786 by mathlove last updated on 22/Dec/21 $${log}\underset{\mathrm{4}{x}} {{x}}+{log}\underset{\frac{{x}}{\mathrm{2}}} {{x}}=\mathrm{2} \\ $$$${solve}\:\:\:{for}\:\:\:{x}=? \\ $$ Answered by mr W last updated on 22/Dec/21 $$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left(\mathrm{4}{x}\right)}+\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left(\frac{{x}}{\mathrm{2}}\right)}=\mathrm{2}…
Question Number 95968 by i jagooll last updated on 29/May/20 $$\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)}\:?\:} \\ $$ Answered by john santu last updated on…
Question Number 161444 by cortano last updated on 18/Dec/21 Commented by blackmamba last updated on 18/Dec/21 $${f}\left({x}\right)=\mathrm{3}{x}−\mathrm{4}\Rightarrow{f}\left(\mathrm{2016}\right)=\mathrm{6044} \\ $$ Commented by blackmamba last updated on…
Question Number 161362 by cortano last updated on 17/Dec/21 $$\:\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…