Solve for x, such that: $sum_{n=0} frac{x}{4^n} = sum_{n=0} frac{1}{x^n}$

Indeed you so have

$${sum_{n=0}^{infty}frac{x}{4^n}=sum_{n=0}^{infty}frac{1}{x^n}=frac{1}{1-frac{1}{x}}}$$

Where the ${frac{1}{1-frac{1}{x}}}$ comes from summing an infinite geometric series with ratio ${frac{1}{x}}$. Note we are also making an implicit assumption here - that ${|x|>1}$ in order for the geometric series to converge.

Assuming the left hand infinite sum converges, dividing through by ${x}$ yields that

$${frac{1}{x - 1} = sum_{n=0}^{infty}frac{1}{4^n}=frac{1}{1-frac{1}{4}}}$$

(again, the rightmost equality comes from summing an infinite geometric series). This implies

$${x-1 = 1-frac{1}{4} = frac{3}{4}}$$

And so

$${x = 1 + frac{3}{4}=frac{7}{4}}$$

As required.

I will refer to the question, since the title has a couple of misleading indices I guess... The LHS is equal to $x (1 + frac{1}{4} + frac{1}{16} + ...) = x frac{1}{1 - frac{1}{4}} = frac{4x}{3}$. The RHS is equal to $frac{1}{1 - frac{1}{x}} = frac{x}{x-1}$, but it converges only for $|x| > 1$. Now we solve $$ frac{4x}{3} = frac{x}{x-1} implies x = 0, frac{7}{4}, $$ but $x = frac{7}{4}$ is the only acceptable solution.

Hint: for $|x|<1$, we have that $$sum_{n=0}^{+infty} x^n=frac{1}{1-x}$$