Question Number 131661 by benjo_mathlover last updated on 07/Feb/21 $$\:\:\:\mathrm{log}\:_{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{2} \\ $$$$\:−\frac{\mathrm{2}\pi}{\mathrm{3}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{3}} \\ $$ Answered by liberty last updated on 07/Feb/21 $$\:\begin{cases}{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\:>\mathrm{0}\:;\:\mathrm{x}\:\mathrm{in}\:\mathrm{I}\:\mathrm{or}\:\mathrm{II}\:\mathrm{quadrant}}\\{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\:\neq\:\mathrm{1}\Rightarrow\mathrm{sin}\:\mathrm{x}\neq\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{1}+\mathrm{cos}\:\mathrm{x}\:=\:\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}}…
Question Number 65972 by gunawan last updated on 07/Aug/19 $$\mathrm{If}\:\frac{\mathrm{log}_{\mathrm{2}} \:{a}}{\mathrm{log}_{\mathrm{3}} \:{b}}={m}\:\mathrm{and}\:\frac{\mathrm{log}_{\mathrm{3}} \:{a}}{\mathrm{log}_{\mathrm{2}} \:{b}}={n} \\ $$$${a}>\mathrm{1}\:\mathrm{and}\:{b}>\mathrm{1} \\ $$$$\mathrm{then}\:\frac{{m}}{{n}}=… \\ $$$${a}.\mathrm{log}_{\mathrm{2}} \:\mathrm{3} \\ $$$${b}.\:\mathrm{log}_{\mathrm{3}} \:\mathrm{2} \\…
Question Number 65970 by gunawan last updated on 07/Aug/19 $$\mathrm{log}_{\mathrm{5}} \sqrt{\mathrm{27}}×\mathrm{log}_{\mathrm{9}} \mathrm{125}+\mathrm{log}_{\mathrm{16}} \mathrm{12}=… \\ $$$${a}.\:\frac{\mathrm{61}}{\mathrm{36}} \\ $$$${b}.\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${c}.\:\frac{\mathrm{61}}{\mathrm{20}} \\ $$$${d}.\:\frac{\mathrm{41}}{\mathrm{12}} \\ $$$${e}.\:\frac{\mathrm{7}}{\mathrm{2}} \\ $$…
Question Number 65971 by gunawan last updated on 07/Aug/19 $$\mathrm{If}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{12}\:\mathrm{then}\:\mathrm{log}\left(\:^{\mathrm{3}} \sqrt{\frac{{b}}{{a}}}\right)=.. \\ $$$${a}.\:−\mathrm{2} \\ $$$${b}.\:−\mathrm{1} \\ $$$${c}.\:\mathrm{0} \\ $$$${d}.\:\mathrm{1} \\ $$$${e}.\:\mathrm{2} \\ $$…