# 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 $$0 everything was as expected and nothing intriguing happened. However, for $$a>1$$ I noticed how $$x^2=a^x$$ had two real solutions for $$1. 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

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 $$e^\frac{2}{e} \approx 2.087065$$

$$x^2=a^x\iff2\frac{\log x}x=\log a$$ has a single solution in $$x$$ when the LHS reaches its unique maximum value, $$\dfrac2e$$, and then

$$a=e^{2/e}=2.0870652286345329598449611070239\cdots$$

We can assume $$a>1$$, because for $$0 we can rewrite the equation as $$(-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 $$\dfrac{2}{\log a} that is, $$a>e^{2/e}\approx 2.08706522863453295984$$ For $$a=e^{2/e}$$ the equation has two solutions; for $$1 the equation has three solutions.

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.