It is natural to take the logarithm, but we do not have to. Here is an informative first step. Let $b=ca$. If we substitute $ca$ for $b$ in our expression, after a small amount of calculation we obtain $\frac{cx+(1-x)}{c^x}.$ This is useful, and not only as a simplifying device: It tells us that only the ratio $b/a$ matters.
Differentiation is straightforward: We get $\frac{c^x(c-1)-(cx+ 1-x)(\ln c)c^x}{c^{2x}}.$ The sign of the derivative is determined by the sign of $(c-1)-(cx+1-x)\ln c.$
Added: Suppose that $a \ne b$. We show that the maximum does not occur at an endpoint. By interchanging the roles of $a$ and $b$ if necessary, we can assume without loss of generality that $c>1$.
Let $g(x)=(c-1)-(cx+1-x)\ln c$. For $x$ positive but very close to $0$, $g(x)\approx -1-\ln c$. But from the Taylor expansion of $\ln(1+t)$, or otherwise, it is easy to see that $\ln(c) when $c-1$ is positive, so $g(x)$ is positive when $x$ is close enough to $0$. A similar argument shows that $g(x)$ is negative when $x$ is close enough to $1$. Thus the maximum occurs in the open interval $(0,1)$. In particular, the unique $x$ at which $(c-1)-(cx+1-x)\ln c=0$ is in the interior of $[0,1]$.