Question Number 164985 by cortano1 last updated on 24/Jan/22 Answered by Eulerian last updated on 24/Jan/22 $$\: \\ $$$$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{3}\centerdot\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)}\:−\:\mathrm{1}\:−\:\mathrm{9}\centerdot\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{log}^{\mathrm{2}} \left(\mathrm{x}\right)}\:=\:\mathrm{25}…
Question Number 164806 by saboorhalimi last updated on 22/Jan/22 Commented by cortano1 last updated on 22/Jan/22 $$\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{3}\right)={a}\:\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{12}\right)=\frac{\mathrm{1}}{{a}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\frac{\mathrm{1}}{{a}}\: \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\:\frac{\mathrm{1}−{a}}{\mathrm{2}{a}}\:…
Question Number 99050 by bemath last updated on 18/Jun/20 $$\mathrm{5}^{\mathrm{5}−\mathrm{3}{x}} \:+\:\mathrm{2}^{{x}+\mathrm{5}} \:=\:\mathrm{5}^{\mathrm{7}−\mathrm{3}{x}} \:−\mathrm{2}^{{x}+\mathrm{6}} \: \\ $$ Commented by bobhans last updated on 18/Jun/20 $$\frac{\mathrm{5}^{\mathrm{5}} }{\mathrm{5}^{\mathrm{3x}}…
Question Number 32885 by scientist last updated on 05/Apr/18 Answered by MJS last updated on 05/Apr/18 $$\mathrm{log}\:{u}={a} \\ $$$$\mathrm{log}\:{v}={b} \\ $$$$\mathrm{log}\:{s}={c} \\ $$$$\mathrm{log}\:{t}={d} \\ $$$$\mathrm{all}\:\mathrm{we}\:\mathrm{need}\:\mathrm{is}\:\mathrm{log}\:\frac{{p}}{{q}}=\mathrm{log}\:{p}−\mathrm{log}\:{q}…
Question Number 97725 by john santu last updated on 09/Jun/20 $$\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{4x}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{x}\right) \\ $$$$\mathrm{find}\:\mathrm{x}\:? \\ $$ Commented by prakash jain last updated on 09/Jun/20…
Question Number 163061 by tounghoungko last updated on 03/Jan/22 $$\:\sqrt[{\mathrm{log}\:_{{x}} \left(\frac{\mathrm{243}}{{x}}\right)}]{\mathrm{2}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{log}\:_{{x}} \left(\frac{{x}^{\mathrm{5}} }{\mathrm{9}}\right)}\: \\ $$ Answered by mahdipoor last updated on 03/Jan/22 $${get}\:\mathrm{5}−\mathrm{2}{log}_{{x}} \mathrm{3}={u} \\…
Question Number 163060 by tounghoungko last updated on 03/Jan/22 $$\:\:\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)}…