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Question Number 99257 by peter frank last updated on 19/Jun/20

Answered by Ar Brandon last updated on 19/Jun/20

$$1\mathrm{a}\mathrm{\setminus }{\log }_{\mathrm{ab}}\mathrm{x}=\frac{{\log }_{\mathrm{x}}\mathrm{x}}{{\log }_{\mathrm{x}}\mathrm{ab}}=\frac{1}{{\log }_{\mathrm{x}}\mathrm{a}+{\log }_{\mathrm{x}}\mathrm{b}}=\frac{1}{\frac{{\log }_{\mathrm{a}}\mathrm{a}}{{\log }_{\mathrm{a}}\mathrm{x}}+\frac{{\log }_{\mathrm{b}}\mathrm{b}}{{\log }_{\mathrm{b}}\mathrm{x}}}$$$$=\frac{1}{\frac{1}{{\log }_{\mathrm{a}}\mathrm{x}}+\frac{1}{{\log }_{\mathrm{b}}\mathrm{x}}}=\frac{1}{\left\{\frac{{\log }_{\mathrm{b}}\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{x}}{\left({\log }_{\mathrm{a}}\mathrm{x}\right)\left({\log }_{\mathrm{b}}\mathrm{x}\right)}\right\}}$$$$=\frac{\left({\log }_{\mathrm{a}}\mathrm{x}\right)\left({\log }_{\mathrm{b}}\mathrm{x}\right)}{{\log }_{\mathrm{a}}\mathrm{x}+{\log }_{\mathrm{b}}\mathrm{x}}$$

Answered by mahdi last updated on 20/Jun/20

$$1){\log }_{\mathrm{ab}}\mathrm{x}=\frac{1}{{\log }_{\mathrm{x}}\mathrm{ab}}=\frac{1}{{1\mathrm{og}}_{\mathrm{x}}\mathrm{a}+{\log }_{\mathrm{x}}\mathrm{b}}=$$$$\frac{1}{\frac{1}{{\log }_{\mathrm{a}}\mathrm{x}}+\frac{1}{{\log }_{\mathrm{b}}\mathrm{x}}}=\frac{{\log }_{\mathrm{b}}\mathrm{x}.{\log }_{\mathrm{a}}\mathrm{x}}{{1\mathrm{og}}_{\mathrm{b}}\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{x}}(\mathrm{not}=\frac{{\log }_{\mathrm{b}}\mathrm{x}-{\log }_{\mathrm{a}}\mathrm{x}}{{\log }_{\mathrm{b}}\mathrm{x}+{\log }_{\mathrm{a}}\mathrm{x}})$$$$2)2\log (\mathrm{a}+\mathrm{b})=\log ({\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab})=$$$$\log \left({\mathrm{a}}^2\times (1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}+\frac{2\mathrm{b}}{\mathrm{a}})\right)=2\mathrm{loga}+\log (1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}+\frac{2\mathrm{b}}{\mathrm{a}})\left(\mathrm{not}{\log }_{\mathrm{c}}\right)$$$$3)={\log }_2\left(\frac{\mathrm{sinx}}{\mathrm{cosx}(1-\mathrm{tanx})(1+\mathrm{tanx})}\right)=$$$${\log }_2\left(\frac{\mathrm{sinx}}{\mathrm{cosx}(1-{\tan }^2\mathrm{x})}\right)={\log }_2\left(\frac{\mathrm{sinx}}{\mathrm{cosx}-\mathrm{tanxsinx}}\right)$$$${\log }_2\left(\frac{\mathrm{sinxcosx}}{{\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}}\right)={\log }_2\left(\frac{\frac{1}{2}\times \sin \left(2\mathrm{x}\right)}{\cos \left(2\mathrm{x}\right)}\right)=$$$${\log }_2\left(\tan \left(2\mathrm{x}\right)\right)-1$$

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