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Category: Logarithms

Question-170669

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}} }…

log-3-x-logx-solve-for-X-

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-170327

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}\:…

A-z-C-2-lt-z-lt-4-fine-log-A-where-log-is-complex-logaritmique-

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-169878

Question Number 169878 by mathlove last updated on 11/May/22 Answered by nikif99 last updated on 11/May/22 $$\frac{\cancel{\mathrm{log}\:\mathrm{3}}}{\mathrm{log}\:\mathrm{2}}\:\centerdot\:\frac{\cancel{\mathrm{log}\:\mathrm{4}}}{\cancel{\mathrm{log}\:\mathrm{3}}}\:\centerdot\:\frac{\cancel{\mathrm{log}\:\mathrm{5}}}{\cancel{\mathrm{log}\:\mathrm{4}}}\:\centerdot\:…\:\centerdot\:\frac{\mathrm{log}\:\left({a}+\mathrm{1}\right)}{\cancel{\mathrm{log}\:{a}}}\:=\:\mathrm{5}\:\Rightarrow \\ $$$$\frac{\mathrm{log}\:\left({a}+\mathrm{1}\right)}{\mathrm{log}\:\mathrm{2}}\:=\:\mathrm{5}\:\Rightarrow\:\mathrm{log}\:\left({a}+\mathrm{1}\right)\:=\:\mathrm{5}\:\mathrm{log}\:\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{log}\:\left({a}+\mathrm{1}\right)\:=\:\mathrm{log}\:\mathrm{2}^{\mathrm{5}} \:\Rightarrow\:{a}+\mathrm{1}=\:\mathrm{2}^{\mathrm{5}} \:\Rightarrow\:{a}=\mathrm{31} \\ $$…

what-is-2-log2x-3-log3x-

Question Number 103961 by MMpotulo last updated on 18/Jul/20 $${what}\:{is}\: \\ $$$$\mathrm{2}^{{log}\mathrm{2}{x}} =\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$ Answered by bramlex last updated on 18/Jul/20 $$\mathrm{ln}\:\left(\mathrm{2}^{\mathrm{ln}\:\left(\mathrm{2}{x}\right)} \right)\:=\:\mathrm{ln}\:\left(\mathrm{3}^{\mathrm{ln}\:\left(\mathrm{3}{x}\right)} \right)…