Question Number 13151 by Tinkutara last updated on 15/May/17 Answered by mrW1 last updated on 16/May/17 $${x}>\mathrm{9} \\ $$$$\mathrm{log}\:\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{2} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{100} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{19}{x}−\mathrm{91}>\mathrm{0} \\…
Question Number 65934 by gunawan last updated on 06/Aug/19 $$\:\mathrm{log}_{\mathrm{7}} \:\mathrm{2}={a} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}={b} \\ $$$$\:\mathrm{log}_{\mathrm{6}} \:\mathrm{98}=… \\ $$ Answered by meme last updated on…
Question Number 65930 by gunawan last updated on 06/Aug/19 $$\mathrm{If}\:{t}=\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{3}{x}+\mathrm{7}}\:\mathrm{then} \\ $$$$\mathrm{log}\left(\mathrm{1}−\mid{t}\mid\right)\:\mathrm{can}\:\mathrm{to}\:\mathrm{find}\:\mathrm{for} \\ $$$$\mathrm{a}.\:\mathrm{2}<{x}<\mathrm{6} \\ $$$${b}.−\:\mathrm{2}<{x}<\mathrm{5} \\ $$$${c}.−\:\mathrm{2}\leqslant{x}\leqslant\mathrm{6} \\ $$$$\mathrm{d}.\:{x}\leqslant−\mathrm{2}\:\mathrm{or}\:{x}>\mathrm{6} \\ $$$${e}.\:{x}<−\mathrm{1}\:{or}\:{x}>\mathrm{3} \\ $$…
Question Number 65931 by gunawan last updated on 06/Aug/19 $$\frac{\left(\mathrm{log}_{\mathrm{6}} \mathrm{36}\right)^{\mathrm{2}} −\left(\mathrm{log}_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\mathrm{log}_{\mathrm{3}} \left(\sqrt{\mathrm{12}}\right)}=… \\ $$ Answered by $@ty@m123 last updated on 06/Aug/19 $$=\frac{\left(\mathrm{log}\:_{\mathrm{6}}…
Question Number 131457 by pete last updated on 04/Feb/21 $$\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} \\…