$\text{ The value of }\underset{x\to a}{\lim }\frac{\log (x-a)}{\log \left(e^x-e^a\right)}\text{ is }$

$\begin{array}{l}\underset{x\to a}{\lim }\frac{\log (x-a)}{\log \left(e^x-e^a\right)} \left(\frac{\infty }{\infty } model\right)\\ =\underset{x\to a}{\lim }\frac{\frac{1}{x-a}}{\frac{1}{e^x-e^a}·\left(e^x-0\right)} \left( by L Hospital rule\right)\\ =\underset{x\to a}{\lim }\frac{1}{x-a}\times \frac{e^x-e^a}{e^x}\\ =\underset{x\to a}{\lim }\frac{1}{e^x}\frac{e^a\left(e^{x-a}-1\right)}{x-a}\\ =\underset{x\to a}{\lim }\frac{e^a}{e^x}\underset{x\to a}{\lim }\frac{e^{x-a}-1}{x-a} \left(\sin ce \underset{x\to 0}{\lim }\frac{e^x-1}{x}=1\right)\\ =\frac{e^a}{e^a·1}\\ =1\end{array}$