ABC26GN2064 · Logarithms

$$\begin{aligned} & \log _{12} 27=a \Rightarrow \frac{\log 27}{\log 12}=a \Rightarrow \frac{\log 3^{3}}{\log \left(3 \times 2^{2}\right)}=a \\ & \Rightarrow \frac{3 \log 3}{\log 3+2 \log 2}=a \Rightarrow \frac{\log 3+2 \log 2}{3 \log 3}=\frac{1}{a} \\ & \Rightarrow \frac{\log 3}{3 \log 3}+\frac{2 \log 2}{3 \log 3}=\frac{1}{a} \Rightarrow \frac{2}{3} \frac{\log 2}{\log 3}=\frac{1}{a}-\frac{1}{3}=\left(\frac{3-a}{3 a}\right) \\ & \Rightarrow \frac{\log 2}{\log 3}=\left(\frac{3-a}{2 a}\right) \Rightarrow \log 3=\left(\frac{2 a}{3-a}\right) \log 2 \\ & \log _{6} 16=\frac{\log 16}{\log 6}=\frac{\log 2^{4}}{\log (2 \times 3)}=\frac{4 \log 2}{\log 2+\log 3}=\frac{4 \log 2}{\log 2\left[1+\left(\frac{2 a}{3-a}\right)\right]} \\ & =\frac{4}{\left(\frac{3+a}{3-a}\right)}=\frac{4(3-a)}{(3+a)} \end{aligned}$$