If $x^2=a^x$ has exactly one root then find $a$ I was playing around in Desmos, and amidst that I became interested in how the graphs of exponentials and quadratics interact. For $0<a<1$ everything was as expected and nothing intriguing happened. However, for $a>1$ I noticed how $x^2=a^x$ had two real solutions for $1<a<2$. After playing around with the slider for a bit more I saw that this was true till about $a=2.1$ For any value of $a$ after that the graphs never met for $x>0$. So, I thought of finding the exact value of $a$ for which the two graphs just touch, or in other words, have exactly one solution.
At first I thought it would be trivial to find out at exactly what value of $a$ this happened. But when I actually sat down to try the problem, I realised I had no idea what to do. At first I thought the derivatives of the two functions would be equal when they touch, and even though that's true I found that the derivatives could be equal at other places too (not to say I have no idea how to solve $2x=a^x\ln a$ either)
I'm sure I am again making a big fuss out of something trivial. But, I just can't seem to see how we can solve this (with elementary maths). Can someone please point me in the right direction? Thanks
 A: Why don't you write it as $\left(x^\frac{1}{x}\right)^2$ and draw its graph?
If I correctly remember, the maxima occurs at $x=e$ (easily verified by differentiating.)
So your value of $a$ for only one solution is $e^\frac{2}{e} \approx 2.087065$
A: $$x^2=a^x\iff2\frac{\log x}x=\log a$$ has a single solution in $x$ when the LHS reaches its unique maximum value, $\dfrac2e$, and then
$$a=e^{2/e}=2.0870652286345329598449611070239\cdots$$
A: We can assume $a>1$, because for $0<a<1$ we can rewrite the equation as $(-x)^2=(a^{-1})^{-x}$. The case $a=1$ is obvious.
Since $0$ is not a solution, we can write the equation as $2\log\lvert x\rvert=x\log a$ and observe that, with
$$
f(x)=2\log\lvert x\rvert-x\log a
$$
we have
$$
\lim_{x\to-\infty}f(x)=\infty,\quad
\lim_{x\to0}f(x)=-\infty,\quad
\lim_{x\to\infty}f(x)=-\infty
$$
Thus there is always a negative solution. If you want a single solution, the maximum over the interval $(0,\infty)$ has to be negative. Since
$$
f'(x)=\dfrac{2}{x}-\log a
$$
the maximum is at $2/\log a$ and
$$
f(2/\log a)=2\log\frac{2}{\log a}-2
$$
which is negative if and only if
$$
\dfrac{2}{\log a}<e
$$
that is,
$$
a>e^{2/e}\approx 2.08706522863453295984
$$
For $a=e^{2/e}$ the equation has two solutions; for $1<a<e^{2/e}$ the equation has three solutions.

A: Your premise is wrong. The equation
$$x^2=a^x$$
it has two roots in the semi-plane $x>0$, namely: $(x,a)=(2,2);(4,2)$. Also, if you plot $x$ vs $a$, for $x>0$, you will find that the resulting curve has the same slope in at least two different places.
