## Task $3 / 65$

The equation $a^{x}=a x$ has for every positive real number $a$ a solution $x_{1}=1$. What condition must the number $a$ satisfy so that the equation

a) has another solution $x_{2} \neq 1$

b) has no solution other than $x_{1}=1$?
The solutions of the equation $a^{x}=a x$ are the zeros of the continuous function $y=a^{x}-a x$, which has the first derivative $y^{\prime} a^{x} \ln a-a$ and the second derivative $y^{\prime \prime}=a^{x}(\ln a)^{2}$.

For every real $x$, $y^{\prime \prime}>0$, so the first derivative is monotonically increasing, and the function has a curve that is concave from above (convex from below), thus having at most one minimum and no maximum.

If the function has no minimum, then $x_{1}=1$ is the only zero. If the function has a minimum and it lies below the $x$-axis, then

$$
x_{2}x_{\min }>x_{1}=1
$$

If the function has a minimum and it lies on the $x$-axis, then $x_{2}=x_{\min}=x_{1}=1$. The minimum cannot lie above the $x$-axis, as the function would then also have a maximum.

To determine the value of $x_{\min }$, we set $y^{\prime}=0$. This yields

$$
x_{\min }=1-\frac{\ln \ln a}{\ln a}
$$

If $01$.

If $a=e$, then $\ln a=1, \ln \ln a=0, x_{\min }=1$, and thus $x_{2}=x_{1}=1$.

If $a>e$, then $\ln a>1, \ln \ln a>0, x_{\text {min }}e$, in case b) $1<a<e$ must hold, and in case c) either $0<a \leq 1$ or $a=e$ must hold.