Find the value of 'x' in a product of exponentiated logarithms Find the value of '$x$' if, $$\large \left(\frac{1}{2^{\log_x 4}} \right) \cdot \left( \frac{1}{2^{\log_x 16}} \right) \cdot \left(\frac{1}{2^{\log_x 256}} \right) \cdots = 2 $$
I tried to make it simple by resulting series is not converging, the suggested answers in my module is either ($2,\frac{1}{2},4,\frac{1}{4}$) 
Also I tried this and this but  not the correct answer in either case.

$$
\begin{align}
\left(\frac{1}{2^{\log_x 4}}\right)\left(\frac{1}{2^{\log_x 16}}\right)\left(\frac{1}{2^{\log_x16}}\right)\cdots&=2\\
2^{-\log_x 4}\times2^{-\log_x 16}\times2^{-\log_x 256}\ldots&=2\\
2^{-\left(\log_x 4+\log_x 16+\log_x 256+\cdots\right)}&=2\\
\log_x(4\times16\times256\times\cdots)&=-1\\
4\times 16\times256\times\cdots&=\frac{1}{x}\\
2^2\times2^4\times2^8\times\cdots&=\frac{1}{x}\\
2^{2+4+8+\cdots}&=\frac{1}{x}
\end{align}
$$
Now say we have, $2+4+8+\cdots$; this can be written as $2+4+8+\cdots+2^n$ when $n\to\infty$. $2+4+8+\cdots+2^n$ is a geometric series and:
$$2+4+8+\cdots+2^n=2\left(2^n-1\right)$$
So now we have:
$$
\begin{align}
x&=\lim_{n\to\infty}\left(\frac{1}{2^{2\left(2^n-1\right)}}\right)
\end{align}$$
