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Question Number 65972 by gunawan last updated on 07/Aug/19

$$\mathrm{If}\frac{{\log }_{2}a}{{\log }_{3}b}=m\mathrm{and}\frac{{\log }_{3}a}{{\log }_{2}b}=n$$$$a>1\mathrm{and}b>1$$$$\mathrm{then}\frac{m}{n}=...$$$$a.{\log }_{2}3$$$$b.{\log }_{3}2$$$$c.{\log }_{4}9$$$$d.{\left({\log }_{2}3\right)}^2$$$$e.{\left({\log }_{3}2\right)}^2$$

Answered by MJS last updated on 07/Aug/19

$$m=\frac{\frac{\ln a}{\ln 2}}{\frac{\ln b}{\ln 3}}=\frac{\ln 3\ln a}{\ln 2\ln b};n=\frac{\frac{\ln a}{\ln 3}}{\frac{\ln b}{\ln 2}}=\frac{\ln 2\ln a}{\ln 3\ln b}$$$$\frac{m}{n}=\frac{\frac{\ln 3\ln a}{\ln 2\ln b}}{\frac{\ln 2\ln a}{\ln 3\ln b}}=\frac{{\left(\ln 3\right)}^2}{{\left(\ln 2\right)}^2}={\left(\frac{\ln 3}{\ln 2}\right)}^2={\left({\log }_{2}3\right)}^2$$

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