Problem source

By considering the maximum of $\ln x-x\ln a$, or otherwise,
show that the equation
$x=a^{x}$ has no real roots if $a > e^{1/e}$.
How many real roots does the equation have if
$0 < a < 1$? Justify your answer.

Solution source

\begin{align*}
&& y &= \ln x - x \ln a \\
\Rightarrow && y' &= \frac1x - \ln a \\
&& y'' &= -\frac{1}{x^2}
\end{align*}
Therefore the maximum is when $x = \frac{1}{\ln a}$ and $y_{max} = -\ln \ln a - 1$. If $y_{max} < 0$ then $y \neq 0$.

But that's equivalent to $a > e^{1/e}$.

\begin{align*}
&& 0 &> -\ln \ln a - 1 \\
\Leftrightarrow && 1 &> - \ln \ln a \\
\Leftrightarrow && \ln \ln a &>-1 \\
\Leftrightarrow && \ln a &> e^{-1} \\
\Leftrightarrow && a & > e^{1/e}
\end{align*}


If $0 < a < 1$ then, when $x$ is small, $\ln x - x \ln a$ is large and negative. When $x$ is large and positive $\ln x$ is positive and $-x \ln a$ is positive. We also notice there is no turning point. Hence exactly one solution