equation with exponential functions

Question

Solve the following equation over the real numbers(preferably without using calculus): $$ 4^x + 4^{1/x} =18 $$

I already know the solutions thanks to Wolfram, what I have trouble with is proving those are the only roots, along with trouble regarding how they are determined.

Answer 1

If $x$ is a solution, $\dfrac1x$ is perforce another. There are no solutions in the negatives (as $f(x)<2$), and we can just discuss $x>1$. As $f(1)<18$ and $f(3)>18$ we are sure that there is a root in this range. By inspection, $2$.

The derivative is

$$\log4\,4^x-\frac{\log4}{x^2}4^{1/x}.$$

It equals $0$ at $x=1$ and remains positive for larger values (the left term increases and the right one decreases), so that the function is monotonic.

The only solutions are $2$ and $\dfrac12$.

Answer 2

It looks like $4$'s or $2$'s are the thing here, so I'd write

$$4^x+4^{1/x} = 2(4-1)^2 = 2(4^2-2\cdot 4 +1) = 2^5-2^4+2.$$

Looking at this, I think $2$'s are the way to go:

$$2^{2x}+2^{2/x} = 2^5-2^4+2 = 2^4 +2.$$

This makes it pretty clear that $2x =4$ and $2x=1$ give solutions.