Let \(\quad \log _b x=\mathrm{P} \quad \log _b a=\mathrm{Q} \quad \log _a x=\mathrm{R}\)

\[

\therefore x=b^{\mathrm{P}} \quad \therefore a=b^{\mathrm{Q}} \quad \therefore x=a^{\mathrm{R}}

\]

Now since \(x=b^{\mathrm{P}}\) and \(x=a^{\mathrm{R}}\)

\[

\therefore b^{\mathrm{P}}=a^{\mathrm{R}}

\]

But since \(a=b^{\mathrm{Q}}\)

\[

\begin{aligned}

& \therefore b^{\mathrm{P}}=\left(b^{\mathrm{Q}}\right)^{\mathrm{R}} \\

& \therefore b^{\mathrm{P}}=b^{\mathrm{QR}} \\

& \therefore \mathrm{P}=\mathrm{QR} \\

& \therefore \log _b x=\left(\log _b a\right)\left(\log _a x\right) \\

& \therefore \frac{\log _b x}{\log _b a}=\log _a x

\end{aligned}

\]