## Task $6 / 90$

What conditions must the real constants $a ; b>0$ satisfy if the function

$$
f(x)=\frac{a^{x}-b^{x}}{a-b}
$$

is to have a maximum for $x_{0} \geq 0$?
Due to the symmetry with respect to $a$ and $b$, it suffices to examine the function for $a > b$. Since a constant factor $\frac{1}{a-b}$ does not affect the existence and location of extremum points, the function $f(x)=\frac{a^{x}-b^{x}}{a-b}$ can be replaced by the function $\varphi(x)=a^{x}-b^{x}$ in our case. A necessary condition for the existence of extremum points at $x_{0}$ is

$$
\varphi^{\prime}\left(x_{0}\right)=a^{x_{0}} \ln a-b^{x_{0}} \ln b=0 \quad \text { thus } \quad x_{0}=\frac{\ln \frac{\ln b}{\ln a}}{\ln a-\ln b}
$$

If now $a>1$ and $a>b$, then (due to the strict monotonicity of the $\ln$-function) $\ln a>\ln b$ and thus $\frac{\ln b}{\ln a}<1$. Since $\ln \frac{\ln b}{\ln a}<0$ and $\ln a - \ln b > 0$, it follows that $x_{0}<0$, which contradicts the requirement of the problem $x_{0} \geq 0$. Therefore, it must be that $0<a ; b<1$.

For $f(x)=\frac{a^{x}-b^{x}}{a-b}$, it follows that a necessary condition for the existence of a maximum is that $0<a ; b<1$.