Question Number 131457 by pete last updated on 04/Feb/21 $$\mathrm{Find}\:{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{3}^{{x}+\mathrm{1}} \:=\:\mathrm{4}^{{x}−\mathrm{1}} \\ $$ Answered by bramlexs22 last updated on 05/Feb/21 $$\Leftrightarrow\:\left({x}+\mathrm{1}\right)\:\mathrm{ln}\:\mathrm{3}\:=\:\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{4} \\ $$$$\Leftrightarrow\:{x}\left(\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{4}\right)\:=\:−\mathrm{ln}\:\mathrm{4}−\mathrm{ln}\:\mathrm{3} \\ $$$$\Leftrightarrow\:{x}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{ln}\:\mathrm{3}\right)=\mathrm{ln}\:\mathrm{4}+\mathrm{ln}\:\mathrm{3}…
Question Number 65781 by gunawan last updated on 03/Aug/19 $$\mathrm{If}\:{xyz}\:\neq\:\mathrm{0}\:\mathrm{and}\:{x}+{y}+{z}=\mathrm{0} \\ $$$${a}=\mathrm{10}^{{z}} \\ $$$${b}=\mathrm{10}^{{y}} \\ $$$${c}=\mathrm{10}^{{x}} \\ $$$$\mathrm{then} \\ $$$${a}^{\left(\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)} .\:{b}^{\left(\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}\right)} .{c}^{\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)} =… \\ $$$${a}.\:\mathrm{0}.\mathrm{001}…
Question Number 223 by 123456 last updated on 25/Jan/15 $$\mathrm{ln}\:\left({e}^{\mathrm{2}} \right)+\mathrm{log}_{\mathrm{10}} \left(\mathrm{100}\right) \\ $$ Answered by ghosea last updated on 16/Dec/14 $$\mathrm{ln}\:{e}^{\mathrm{2}} +\mathrm{log}_{\mathrm{10}} \mathrm{10}^{\mathrm{2}} =\mathrm{2}+\mathrm{2}=\mathrm{4}…
Question Number 143812 by liberty last updated on 18/Jun/21 $$\:\mathrm{log}\:_{\mathrm{a}} \left(\mathrm{ax}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{ax}\right)=\mathrm{log}\:_{\mathrm{a}^{\mathrm{2}} } \left(\frac{\mathrm{1}}{\mathrm{a}}\right) \\ $$$$\:\mathrm{a}>\mathrm{0}\:,\:\mathrm{a}\neq\mathrm{1}\:.\:\mathrm{So}\:\mathrm{x}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 18/Jun/21…
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}\:? \\…