
p.s I have no idea how to type math on this program so I just copied and pasted from a document.
p.p.s I also tried doing it with log base a and that was a travesty. This was my best attempt because it was the nicest looking

p.s I have no idea how to type math on this program so I just copied and pasted from a document.
p.p.s I also tried doing it with log base a and that was a travesty. This was my best attempt because it was the nicest looking
First observe that $x\le 0$ can't be a solution since $2a>2$ on the right side.
Then, use arithmetic and geometric mean for $xa^{1/x}$ and $a^x/x$. We obtain $$2a^{\frac{x+\frac1x}2}\le xa^{\frac1x}+\frac{a^x}x$$ Now use that $2\le x+\displaystyle\frac1x$ (again by A-G means) and that $a>1$ so we have $a=a^1\le \displaystyle a^{\frac{x+\frac1x}2}$, finally, $$2a\le 2a^{\frac{x+\frac1x}2}\le xa^{\frac1x}+\frac{a^x}x $$ and equality can hold only if $x=1$ by the last step.
Multiplying by $x$ and rearranging we get:
$$x^2a^{1/x} -2ax + a^x = 0.$$
From the quadratic formula,
$$x = \frac{2a \pm \sqrt{4a^2 - 4a^{x + (1/x)}}}{2a^{1/x}}.$$
Now, $x$ must be positive, because $a$ is positive, and the square root is always less than $2a$.
However, the exponent under the square root, $x + \frac{1}{x}$ must be less than or equal to $2$ in order to keep the value under the square root positive.
This happens only at $x=1.$