Question

Prove that \(\log _a x=\frac{\log _b x}{\log _b a}\)

Answer 1

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 \(a=b^{\mathrm{Q}}\)
\[
\begin{aligned}
& \therefore b^{\mathrm{P}}=\left(b^{\mathrm{Q}}\right)^{\mathrm{R}} \\
& \therefore b^{\mathrm{P}}=b^{\mathrm{QR}} \\
& \therefore \mathrm{P}=\mathrm{QR} \\
& \therefore \log _b x=\left(\log _b a\right)\left(\log _a x\right) \\
& \therefore \frac{\log _b x}{\log _b a}=\log _a x
\end{aligned}
\]