Since we have a $\frac{0}{0}$ form, we can apply the L'Hôpital's rule.
$$\begin{align}
L &= \lim_{x \to 0}\frac{2^x-1}{x}\\[1em]
&= \lim_{x \to 0}\frac{\frac{d}{dx}(2^x-1)}{\frac{d}{dx}x}\\[1em]
&= \lim_{x \to 0}\frac{2^x\ln 2}{1}\\[1em]
&= \ln 2\\[1em]
L &= \log_e{(2)}
\end{align}$$
Hence, option c is correct.