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Question Number 170669 by solomonwells last updated on 28/May/22 Commented by kaivan.ahmadi last updated on 28/May/22 $${y}={log}_{\mathrm{6}} {x}\Rightarrow{x}=\mathrm{6}^{{y}} \\ $$$$\mathrm{6}^{{y}^{\mathrm{2}} } +\left(\mathrm{6}^{{y}} \right)^{{y}} =\mathrm{12}\Rightarrow\mathrm{2}×\mathrm{6}^{{y}^{\mathrm{2}} }…
Question Number 170661 by solomonwells last updated on 28/May/22 $$\boldsymbol{\mathrm{log}}\:^{\mathrm{3}} \sqrt{}\boldsymbol{\mathrm{x}}\:\:\:=\:\:\:\sqrt{\boldsymbol{\mathrm{logx}}} \\ $$$$\boldsymbol{\mathrm{solve}}\:\:\boldsymbol{\mathrm{for}}\:\:\:\:\:\:\boldsymbol{{X}} \\ $$$$ \\ $$ Commented by kaivan.ahmadi last updated on 28/May/22 $$\frac{\mathrm{1}}{\mathrm{3}}{logx}=\left({logx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 104835 by bemath last updated on 24/Jul/20 $$\begin{cases}{\mathrm{log}\:_{{p}} \left({q}\right)\:=\:{x}^{\mathrm{2}} }\\{\mathrm{log}\:_{{q}} \left({p}^{\mathrm{3}} \right)\:=\:{x}\:}\end{cases} \\ $$$${find}\:{x}\: \\ $$ Answered by john santu last updated on…
Question Number 170327 by mathlove last updated on 21/May/22 Answered by greougoury555 last updated on 21/May/22 $$\frac{\mathrm{1}}{\mathrm{log}\:_{{a}} \left({abc}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{log}\:_{{b}} \left({abc}\right)}\:+\frac{\mathrm{1}}{\mathrm{log}\:_{{c}} \left({abc}\right)}\:=\: \\ $$$$\mathrm{log}\:_{{abc}} {a}+\mathrm{log}_{{abc}} {b}+\mathrm{log}\:_{{abc}} {c}\:=\:\mathrm{1}\:…
Question Number 104624 by dw last updated on 22/Jul/20 Commented by dw last updated on 22/Jul/20 $$? \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 170020 by cortano1 last updated on 14/May/22 $$\:\:\:\:\mathrm{log}\:_{\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)} \left(\mathrm{2}\right)\:<\:\mathrm{log}\:_{{x}} \left(\mathrm{4}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 169918 by pticantor last updated on 12/May/22 $${A}=\left\{\boldsymbol{{z}}\in\mathbb{C}:\:\mathrm{2}<\mid\boldsymbol{{z}}\mid<\mathrm{4}\right\} \\ $$$$\boldsymbol{{fine}}\:\boldsymbol{{log}}\left(\boldsymbol{{A}}\right) \\ $$$$\boldsymbol{{where}}\:\boldsymbol{{log}}\:\boldsymbol{{is}}\:\boldsymbol{{complex}}\:\boldsymbol{{logaritmique}} \\ $$ Answered by pticantor last updated on 12/May/22 $$\boldsymbol{{besoin}}\:\boldsymbol{{d}}'\boldsymbol{{aide}}\:\boldsymbol{{please}}!! \\…
Question Number 169878 by mathlove last updated on 11/May/22 Answered by nikif99 last updated on 11/May/22 $$\frac{\cancel{\mathrm{log}\:\mathrm{3}}}{\mathrm{log}\:\mathrm{2}}\:\centerdot\:\frac{\cancel{\mathrm{log}\:\mathrm{4}}}{\cancel{\mathrm{log}\:\mathrm{3}}}\:\centerdot\:\frac{\cancel{\mathrm{log}\:\mathrm{5}}}{\cancel{\mathrm{log}\:\mathrm{4}}}\:\centerdot\:…\:\centerdot\:\frac{\mathrm{log}\:\left({a}+\mathrm{1}\right)}{\cancel{\mathrm{log}\:{a}}}\:=\:\mathrm{5}\:\Rightarrow \\ $$$$\frac{\mathrm{log}\:\left({a}+\mathrm{1}\right)}{\mathrm{log}\:\mathrm{2}}\:=\:\mathrm{5}\:\Rightarrow\:\mathrm{log}\:\left({a}+\mathrm{1}\right)\:=\:\mathrm{5}\:\mathrm{log}\:\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{log}\:\left({a}+\mathrm{1}\right)\:=\:\mathrm{log}\:\mathrm{2}^{\mathrm{5}} \:\Rightarrow\:{a}+\mathrm{1}=\:\mathrm{2}^{\mathrm{5}} \:\Rightarrow\:{a}=\mathrm{31} \\ $$…
Question Number 169798 by mathlove last updated on 09/May/22 $${log}_{{e}} \left({e}^{\mathrm{2}} {x}^{{lnx}} \right)={log}_{{e}} \left({x}^{\mathrm{3}} \right) \\ $$$${faind}\:\:{x}=? \\ $$ Commented by cortano1 last updated on…