Explanation

(b)
$\lim_{x \rightarrow a}\frac{\log{(x - a)}}{\log{(e^{x} - e^{a})}} = \lim_{x \rightarrow a}\frac{\frac{1}{x - a}}{\frac{e^{x}}{e^{x} - e^{a}}}$
[by L’ Hospital’s rule]
$= \lim_{x \rightarrow a}\ \frac{e^{x} - e^{a}}{e^{x}(x - a)}$
$= \lim_{x \rightarrow a}\ \ \frac{e^{x}}{e^{x}(x - a) + e^{x}}$
[by L’ Hospital’s rule]
$= \frac{e^{a}}{0 + e^{a}} = 1$