Question Number 35899 by imamu222 last updated on 25/May/18 $${Evaluate}\:{log}_{\sqrt{\mathrm{2}}} \mathrm{4}+{log}_{\mathrm{1}/\mathrm{2}} \mathrm{16}−{log}_{\mathrm{4}} \mathrm{32} \\ $$ Commented by Cheyboy last updated on 26/May/18 $${log}_{\sqrt{\mathrm{2}}} \mathrm{4}+{log}_{\mathrm{1}/\mathrm{2}} \mathrm{16}−{log}_{\mathrm{4}}…
Question Number 35897 by imamu222 last updated on 25/May/18 $${log}_{{x}^{\mathrm{1}/\mathrm{2}\:} } \mathrm{64}=\mathrm{3}.\:{What}\:{is}\:{x}? \\ $$ Commented by Cheyboy last updated on 25/May/18 $${log}_{{x}^{\mathrm{1}/\mathrm{2}} } \mathrm{64}=\mathrm{3} \\…
Question Number 166687 by cortano1 last updated on 25/Feb/22 $$\:\:\:\mathrm{log}\:_{\left(\mathrm{2x}−\mathrm{1}\right)} \left(\mathrm{x}+\mathrm{1}\right)\:>\:\mathrm{log}\:_{\left(\mathrm{4}−\mathrm{2x}\right)} \left(\mathrm{x}+\mathrm{1}\right) \\ $$$$\:\:\mathrm{x}=? \\ $$ Commented by Nimatullah last updated on 25/Feb/22 $${p} \\…
Question Number 100832 by student work last updated on 28/Jun/20 $$\mathrm{log}\sqrt{\mathrm{125}}\:\centerdot\mathrm{ln10}\:\centerdot\mathrm{log}_{\mathrm{5}} \mathrm{e}=? \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$ Commented by student work last updated on 28/Jun/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{practice}\:\mathrm{sir}?…
Question Number 100720 by bemath last updated on 28/Jun/20 Commented by 1549442205 last updated on 28/Jun/20 $$ \\ $$$$\mathrm{one}\:\mathrm{other}\:\mathrm{way}:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}:\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}>\mathrm{0}\Leftrightarrow\mathrm{x}>\mathrm{5} \\ $$$$\mathrm{Since}\:\mathrm{log}_{\mathrm{8}} \frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}=\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{x}−\mathrm{5}}{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 100675 by bobhans last updated on 28/Jun/20 $$\mathrm{If}\:\mathrm{log}\:_{\mathrm{2x}} \left(\frac{\mathrm{1}}{\mathrm{18}}\right)\:=\:\mathrm{log}\:_{\mathrm{18}} \left(\frac{\mathrm{1}}{\mathrm{3y}}\right)\:=\:\mathrm{log}\:_{\mathrm{3y}} \left(\frac{\mathrm{1}}{\mathrm{2x}}\right) \\ $$$$\mathrm{find}\:\mathrm{3x}−\mathrm{2y}\: \\ $$ Commented by bramlex last updated on 28/Jun/20 $$\Leftrightarrow\:\mathrm{2}{x}\:=\:\mathrm{3}{y}\:=\:\mathrm{18}\:\rightarrow\begin{cases}{{x}=\mathrm{9}}\\{{y}=\mathrm{6}}\end{cases}…
Question Number 100666 by bobhans last updated on 28/Jun/20 $$\mathrm{find}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequality} \\ $$$$\left(\mathrm{log}\:_{\mathrm{2}} {x}\:−\mathrm{2}\right)^{\mathrm{3}{x}−\mathrm{1}} \:<\:\left(\mathrm{log}\:_{\mathrm{2}} {x}−\mathrm{2}\right)^{\mathrm{3}−{x}} \\ $$ Commented by Rasheed.Sindhi last updated on 28/Jun/20 $$\left(\mathrm{log}\:_{\mathrm{2}}…
Question Number 165818 by MikeH last updated on 08/Feb/22 $$\mathrm{A}\:\mathrm{uniform}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{weight}\:{W} \\ $$$$\mathrm{rest}\:\mathrm{between}\:\mathrm{a}\:\mathrm{smooth}\:\:\mathrm{vertical} \\ $$$$\mathrm{plane}\:\mathrm{and}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{plane}\:\mathrm{inclined} \\ $$$$\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the}\:\mathrm{vertical} \\ $$$$\mathrm{plane}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{reaction}\:\mathrm{at}\:\mathrm{the}\: \\ $$$$\mathrm{contact}\:\mathrm{surfaces}.\: \\ $$ Answered by ajfour…
Question Number 100151 by student work last updated on 25/Jun/20 $$\sqrt{\mathrm{lnx}}\:−\mathrm{ln}\sqrt{\mathrm{x}}\:=\mathrm{0}\:\:\:\:\:\:\mathrm{x}=? \\ $$ Commented by bobhans last updated on 25/Jun/20 $$\mathrm{set}\:\sqrt{\mathrm{ln}\left(\mathrm{x}\right)}\:=\:\mathrm{u} \\ $$$$\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}\left(\mathrm{2}−\mathrm{u}\right)=\mathrm{0} \\…
Question Number 100150 by student work last updated on 25/Jun/20 $$\mathrm{x}^{\mathrm{ln}} −\mathrm{e}^{\mathrm{6}} \centerdot\mathrm{x}=\mathrm{0}\:\:\:\:\:\:\mathrm{x}=? \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$ Commented by student work last updated on 25/Jun/20…