Let the function $$f(x) = \begin{cases} \frac{\log_e(1+5x) - \log_e(1+\alpha x)}{x} & \text{if } x \neq 0 \\ 10 & \text{if } x = 0 \end{cases}$$ be continuous at $$x = 0$$. Then $$\alpha$$ is equal to

We need the function $$f(x) = \begin{cases} \dfrac{\log_e(1+5x) - \log_e(1+\alpha x)}{x} & \text{if } x \neq 0 \\ 10 & \text{if } x = 0 \end{cases}$$ to be continuous at $$x = 0$$.

For continuity, we need $$\displaystyle\lim_{x \to 0} f(x) = f(0) = 10$$. We compute the limit using the standard result $$\displaystyle\lim_{x \to 0} \dfrac{\log_e(1+kx)}{x} = k$$.

We have $$\displaystyle\lim_{x \to 0} \dfrac{\log_e(1+5x) - \log_e(1+\alpha x)}{x} = \lim_{x \to 0} \dfrac{\log_e(1+5x)}{x} - \lim_{x \to 0} \dfrac{\log_e(1+\alpha x)}{x} = 5 - \alpha$$.

Setting $$5 - \alpha = 10$$, we get $$\alpha = -5$$.

Hence, the correct answer is Option D: $$-5$$.