Question Number 161342 by bobhans last updated on 16/Dec/21 $$\:\:\mathrm{2log}\:_{\mathrm{x}} \left(\mathrm{3}\right)\:\mathrm{log}\:_{\mathrm{3x}} \left(\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{9}\sqrt{\mathrm{x}}} \left(\mathrm{3}\right) \\ $$$$\:\mathrm{x}=? \\ $$ Answered by 1549442205PVT last updated on 16/Dec/21 $${the}\:{given}\:{equation}\:{is}\:{equivalent}\:{to}…
Question Number 95485 by O Predador last updated on 25/May/20 Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\mathrm{2llog}_{\mathrm{ln}\left(\pi\right)} \frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{ln}\left(\pi\right)}}\:−\mathrm{2log}_{\mathrm{ln}\left(\pi\right)} \frac{\mathrm{1}}{\mathrm{x}−\mathrm{lnx}}\:=\:\mathrm{1} \\ $$$$\mathrm{2log}_{\mathrm{ln}\left(\pi\right)} \frac{\mathrm{x}−\mathrm{ln}\pi}{\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{ln}\left(\pi\right)}}\:=\:\mathrm{1}…
Question Number 160987 by cortano last updated on 10/Dec/21 $$\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right)} \left(\mathrm{64}\right)\:=\:\mathrm{6}\: \\ $$ Answered by bobhans last updated on 10/Dec/21 $$\:\:\mathrm{log}\:_{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right)}…
Question Number 160931 by blackmamba last updated on 09/Dec/21 $$\:\:{x}\:=\:\mathrm{2}^{\mathrm{log}\:_{\mathrm{5}} \left({x}+\mathrm{3}\right)} \:;\:{x}=? \\ $$ Answered by cortano last updated on 09/Dec/21 Terms of Service Privacy…
Question Number 95108 by bobhans last updated on 23/May/20 $$\mid\mathrm{1}−\mathrm{log}\:_{\left(\frac{\mathrm{1}}{\mathrm{6}}\right)} \left(\mathrm{x}\right)\mid\:+\mathrm{2}\:=\:\mid\mathrm{3}\:−\mathrm{log}\:_{\left(\frac{\mathrm{1}}{\mathrm{6}}\right)} \left(\mathrm{3}\right)\mid\: \\ $$ Answered by john santu last updated on 23/May/20 $$\mathrm{let}\:\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{x}\right)\:=\:\mathrm{t}\: \\…
Question Number 160426 by alf123 last updated on 29/Nov/21 Commented by mr W last updated on 29/Nov/21 $${one}\:{equation}\:{for}\:{two}\:{unknowns} \\ $$$$\Rightarrow{no}\:{unique}\:{solution}! \\ $$ Commented by Rasheed.Sindhi…
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Question Number 159762 by cortano last updated on 21/Nov/21 $$\:\:\:{Given}\:\mathrm{log}\:_{\mathrm{3}} \left({n}\right)=\:\mathrm{log}\:_{\mathrm{6}} \left({m}\right)=\mathrm{log}\:_{\mathrm{12}} \left({m}+{n}\right) \\ $$$$\:\:\:\frac{{m}}{{n}}\:=\:? \\ $$ Answered by bobhans last updated on 21/Nov/21 $$\:\mathrm{log}\:_{\mathrm{3}}…
Question Number 159639 by bobhans last updated on 19/Nov/21 $$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}−\mathrm{3}^{\mathrm{x}} \right)−\mathrm{3}}\:\leqslant\:\mathrm{1}\: \\ $$ Answered by tounghoungko last updated on 19/Nov/21 $$\:\:\:\frac{{x}−\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}−\mathrm{3}^{{x}} \right)−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{27}\right)}\:\leqslant\:\mathrm{1}\:;\:\mathrm{9}−\mathrm{3}^{{x}}…
Question Number 159507 by cortano last updated on 18/Nov/21 Commented by tounghoungko last updated on 18/Nov/21 $$\:\frac{\mathrm{5}.\mathrm{5}^{−\mathrm{2}{x}} +\mathrm{5}^{\mathrm{3}} .\mathrm{5}^{−\mathrm{2}{x}} }{\frac{\mathrm{5}}{\mathrm{8}}.\mathrm{2}^{\mathrm{2}{x}} }\:=\:\mathrm{4} \\ $$$$\Leftrightarrow\:\mathrm{5}^{−\mathrm{2}{x}} +\mathrm{5}^{\mathrm{2}} .\mathrm{5}^{−\mathrm{2}{x}}…