For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x \to 0} f(x) = f(0)$. In this case, $f(0) = k$. So we need to find the limit of $f(x)$ as $x$ approaches 0.
We need to evaluate $\lim_{x \to 0} \frac{\log(1+ax) + \log(1-bx)}{x}$. Since this is of the form $\frac{0}{0}$, we can use L'Hôpital's Rule.
Differentiating the numerator and denominator with respect to $x$, we get: $$\lim_{x \to 0} \frac{\frac{d}{dx} (\log(1+ax) + \log(1-bx))}{\frac{d}{dx} (x)} = \lim_{x \to 0} \frac{\frac{a}{1+ax} - \frac{b}{1-bx}}{1}$$
Now, we can directly substitute $x=0$ into the expression: $$\lim_{x \to 0} \frac{\frac{a}{1+ax} - \frac{b}{1-bx}}{1} = \frac{a}{1+a(0)} - \frac{b}{1-b(0)} = a - b$$
Since $f(x)$ is continuous at $x=0$, we have $k = \lim_{x \to 0} f(x)$. Therefore, $k = a - b$.
Final Answer: \(a-b\)
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