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…
Question Number 34301 by math1967 last updated on 03/May/18 $${If}\:{log}_{\mathrm{12}} \mathrm{18}={a}\:,{find}\:{log}_{\mathrm{24}} \mathrm{16}\:{in}\:{term} \\ $$$${of}\:\:{a} \\ $$ Answered by MJS last updated on 03/May/18 $$\mathrm{12}^{{a}} =\mathrm{18}…
Question Number 99681 by Algoritm last updated on 22/Jun/20 Answered by floor(10²Eta[1]) last updated on 22/Jun/20 $$\mathrm{1}−\mathrm{2}{x}>\mathrm{0}\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{x}\neq\mathrm{1}\Rightarrow{x}\neq\mathrm{0} \\ $$$${but}\:{we}\:{know}\:{that}\:{for}\:{all}\:{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}−\mathrm{2018}\:{will}\:{be}\:{negative},\:{so}\:{for}\:{that} \\ $$$${expression}\:{be}\:{positive}…
Question Number 165178 by cortano1 last updated on 27/Jan/22 $$\:\:{x}\in{R}\:\Rightarrow\:\mid\:\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\mid^{\mathrm{3}} +\mid\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{2}{x}\right)\mid^{\mathrm{3}} =\mathrm{28} \\ $$ Answered by aleks041103 last updated on 27/Jan/22 $${log}_{\mathrm{2}} \left({x}/\mathrm{2}\right)={log}_{\mathrm{2}}…
Question Number 165136 by saboorhalimi last updated on 26/Jan/22 Answered by MJS_new last updated on 26/Jan/22 $$\mathrm{obviously}\:{a}=\mathrm{8} \\ $$$$\mathrm{log}_{\mathrm{3}} \:\mathrm{9}\:=\mathrm{log}_{\mathrm{4}} \:\mathrm{16}\:=\mathrm{2} \\ $$ Commented by…
Question Number 165068 by cortano1 last updated on 25/Jan/22 Commented by bobhans last updated on 25/Jan/22 $$\:\Rightarrow\mathrm{x}^{\mathrm{log}\:_{\mathrm{81}} \left(\mathrm{243}\right)} −\mathrm{2x}\:=\:\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{log}\:_{\mathrm{16}} \left(\mathrm{2}\right)} −\mathrm{8} \\ $$$$\Rightarrow\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{4}}} −\mathrm{2x}\:=\:\left(\mathrm{x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{8}…