Question Number 121138 by Ar Brandon last updated on 05/Nov/20 Answered by TANMAY PANACEA last updated on 05/Nov/20 $${x}^{{x}−{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} } =\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{{x}+{x}^{\frac{−\mathrm{1}}{\mathrm{2}}} } \\ $$$${x}−{x}^{−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 121100 by Ar Brandon last updated on 05/Nov/20 Answered by TANMAY PANACEA last updated on 05/Nov/20 $$\frac{{logx}}{{x}\left({y}+{z}−{x}\right)}=\frac{{logy}}{{y}\left({z}+{x}−{y}\right)}=\frac{{logz}}{{z}\left({x}+{y}−{z}\right)}={k} \\ $$$${logx}={kx}\left({y}+{z}−{x}\right) \\ $$$${logy}={ky}\left({z}+{x}−{y}\right) \\ $$$${logz}={kz}\left({x}+{y}−{z}\right)…
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Question Number 121002 by Ar Brandon last updated on 04/Nov/20 Answered by MJS_new last updated on 04/Nov/20 $$ \\ $$ Commented by Ar Brandon last…
Question Number 120927 by Ar Brandon last updated on 04/Nov/20 Answered by 675480065 last updated on 04/Nov/20 $$\mathrm{domain}:\:\mathrm{2x}−\mathrm{3}/\mathrm{4}\:>\mathrm{0}\:\cap\:\mathrm{x}>\mathrm{0}\:\cap\:\mathrm{x}>\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{x}>\mathrm{3}/\mathrm{8}\:\cap\:\mathrm{x}>\mathrm{0} \\ $$$$\mathrm{hence}\:\mathrm{2x}−\mathrm{3}/\mathrm{4}\:>\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{3}/\mathrm{4}\:<\mathrm{0}…
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Question Number 54875 by shaddie last updated on 13/Feb/19 $$\mathrm{Given}\:\mathrm{that}\frac{\mathrm{log}\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2x}−\mathrm{1}} }{\mathrm{log}\left(\mathrm{3x}+\mathrm{1}\right)}=\mathrm{5},\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$ Answered by kaivan.ahmadi last updated on 14/Feb/19 $$\mathrm{log}\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2x}−\mathrm{1}} =\mathrm{log}\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{5}} \Rightarrow \\ $$$$\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2x}−\mathrm{1}}…
Question Number 120380 by bramlexs22 last updated on 31/Oct/20 $${Given}\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{10}\right)=\frac{{a}}{{a}−\mathrm{1}}}\\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)=\frac{\mathrm{1}}{{b}}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{1}+\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{15}\right)\:? \\ $$ Answered by FelipeLz last updated on 31/Oct/20 $$\mathrm{log}_{\mathrm{2}}…
Question Number 120120 by benjo_mathlover last updated on 29/Oct/20 $$\begin{cases}{\mathrm{log}\:_{{a}} \left({x}\right)\:=\:\mathrm{8}}\\{\mathrm{log}\:_{{b}} \left({x}\right)\:=\:\mathrm{3}\:}\\{\mathrm{log}\:_{{c}} \left({x}\right)\:=\:\mathrm{6}}\end{cases}\:\Rightarrow\:\mathrm{log}\:_{{abc}} \:\left({x}\right)=? \\ $$ Answered by bemath last updated on 29/Oct/20 $$\begin{cases}{\mathrm{log}\:_{{a}} \left({x}\right)=\mathrm{8}\Rightarrow\mathrm{log}\:_{{x}}…
Question Number 119802 by bemath last updated on 27/Oct/20 $${Given}\:{a},{b},{c}\:{real}\:{number}\:{and}\:{not}\:{equal}\:{to}\:\mathrm{1}. \\ $$$${If}\:\mathrm{log}\:_{{a}} \left({b}\right)+\mathrm{log}\:_{{b}} \left({c}\right)+\mathrm{log}\:_{{c}} \left({a}\right)=\mathrm{0}\:{then}\: \\ $$$$\left(\mathrm{log}\:_{{a}} \left({b}\right)\right)^{\mathrm{3}} +\left(\mathrm{log}\:_{{b}} \left({c}\right)\right)^{\mathrm{3}} +\left(\mathrm{log}\:_{{c}} \left({a}\right)\right)^{\mathrm{3}} =? \\ $$…