ABC26GN2006 · Logarithms

(i) $\log_{5}3 \times \log_{27}25=\frac{\log 3}{\log 5} \times \frac{\log 25}{\log 27}=\frac{\log 3}{\log 5} \times \frac{\log(5^{2})}{\log(3^{3})}=\frac{\log 3}{\log 5} \times \frac{2\log 5}{3\log 3}=\frac{2}{3}$. (ii) Let $\log_{9}27=n$. Then, $9^{n}=27 \Leftrightarrow 3^{2n}=3^{3} \Leftrightarrow 2n=3 \Leftrightarrow n=\frac{3}{2}$. Again, let $\log_{27}9=m$. Then, $27^{m}=9 \Leftrightarrow 3^{3m}=3^{2} \Leftrightarrow 3m=2 \Leftrightarrow m=\frac{2}{3}$. $$\log_{9}27-\log_{27}9=(n-m)=\left(\frac{3}{2}-\frac{2}{3}\right)=\frac{5}{6}.$$