Question Number 143811 by liberty last updated on 18/Jun/21 $$\:\begin{cases}{\mathrm{5}\left(\mathrm{log}\:_{\mathrm{y}} \left(\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{y}\right)\right)=\mathrm{26}}\\{\:\mathrm{xy}\:=\:\mathrm{64}}\end{cases}\mathrm{then} \\ $$$$\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{xy}\:=? \\ $$ Answered by liberty last updated on 18/Jun/21…
Question Number 12489 by Joel577 last updated on 23/Apr/17 $$\mathrm{log}_{\mathrm{2}} \:{x}\:+\:\mathrm{log}_{\mathrm{1}/{x}} \:\frac{\mathrm{1}}{\mathrm{2}}\:\geqslant\:\mathrm{0} \\ $$$${x}\:=\:? \\ $$ Answered by mrW1 last updated on 23/Apr/17 $$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}}+\frac{\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{ln}\:\left(\frac{\mathrm{1}}{{x}}\right)}\geqslant\mathrm{0} \\…
Question Number 143475 by Ghaniy last updated on 14/Jun/21 $$\:{evaluate}; \\ $$$$\frac{\left(\sqrt{\mathrm{7}}\right)^{\mathrm{log64}} −\left(\mathrm{3}\right)^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} }{\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{8}−\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{64}\right)\left(\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{64}}\right)}\right)} \\ $$ Commented by amin96 last updated…
Question Number 12284 by tawa last updated on 17/Apr/17 Answered by mrW1 last updated on 18/Apr/17 $${let}\:{t}=\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} +\mathrm{125}\right)=\mathrm{log}_{\mathrm{5}} \:\mathrm{6}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\:\Rightarrow \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\mathrm{2}{t}}…
Question Number 77805 by Dah Solu Tion last updated on 10/Jan/20 $${Solve}\:{for}\:{x} \\ $$$$\frac{\mathrm{8}^{{x}} \:+\:\mathrm{27}^{{x}} }{\mathrm{12}^{{x}} \:+\:\mathrm{18}^{{x}} }\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$ Answered by john santu last…
Question Number 142755 by ERA last updated on 05/Jun/21 $$\mathrm{log}_{\mathrm{3}} \mathrm{x}^{\mathrm{3}} +\mathrm{log}_{\mathrm{2}} \mathrm{x}^{\mathrm{2}} =\frac{\mathrm{2lg6}}{\mathrm{lg2}}+\mathrm{1}\:\:\mathrm{find}\:\:\mathrm{x} \\ $$ Commented by iloveisrael last updated on 05/Jun/21 $${what}\:{different}\:\mathrm{log}\:\:{and}\:\mathrm{lg}\:? \\…
Question Number 11417 by Joel576 last updated on 25/Mar/17 $${A}\:=\:\mathrm{log}\:\left(\mathrm{5}{x}\:+\:\mathrm{1}\right)\left(\mathrm{3}{x}\:+\:\mathrm{5}\right) \\ $$$${B}\:=\:\frac{\mathrm{1}}{\mathrm{log}\:\left(\mathrm{5}{x}+\:\mathrm{1}\right)\left({x}\:−\:\mathrm{1}\right)} \\ $$$$\mathrm{If}\:{A}\:+\:{B}\:\geqslant\:\mathrm{1},\:{x}\:\mathrm{must}\:\mathrm{be}\:… \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 11383 by tawa last updated on 22/Mar/17 $$\mathrm{There}\:\mathrm{are}\:\mathrm{six}\:\mathrm{trains}\:\mathrm{travelling}\:\mathrm{between}\:\mathrm{Abuja}\:\mathrm{and}\:\mathrm{Lagos}\:\mathrm{and}\:\mathrm{back}. \\ $$$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{a}\:\mathrm{man}\:\mathrm{travel}\:\mathrm{from}\:\mathrm{abuja}\:\mathrm{to}\:\mathrm{Lagos}\:\mathrm{by}\:\mathrm{one}\:\mathrm{train}\:\mathrm{and} \\ $$$$\mathrm{return}\:\mathrm{by}\:\mathrm{a}\:\mathrm{different}\:\mathrm{train} \\ $$ Answered by sandy_suhendra last updated on 23/Mar/17 $$\mathrm{Abuja}\:\mathrm{to}\:\mathrm{Lagos}\:=\:\mathrm{6}\:\mathrm{trains} \\…
Question Number 76034 by hmamarques1994@gmail.com last updated on 22/Dec/19 $$\: \\ $$$$\:\mathrm{53}^{\boldsymbol{\mathrm{log}}_{\boldsymbol{\mathrm{x}}} \left(\mathrm{7}\right)} \:=\:\sqrt{\boldsymbol{\mathrm{x}}} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\: \\ $$ Answered by MJS…
Question Number 141365 by bemath last updated on 18/May/21 $$\:\mathrm{log}\:_{\frac{\mathrm{9}}{\mathrm{4}}} \left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{…}}}}\right)\:=? \\ $$ Answered by EDWIN88 last updated on 18/May/21 $$\mathrm{let}\:\mathcal{E}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\ldots}}}} \\ $$$$\Rightarrow\:\mathcal{E}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\sqrt{\mathrm{6}−\mathcal{E}}\: \\ $$$$\Rightarrow\:\mathcal{E}^{\mathrm{2}}…