Never-ending logarithmic series

Assuming that there should also be the term $\log_{2}(\sqrt{x})$ after $\log_{2}(x)$ then the equation becomes

$\log_{2}(x) + \log_{2}(x^{\frac{1}{2}}) + \log_{2}(x^{\frac{1}{4}}) + \log_{2}(x^{\frac{1}{8}}) + ...... = 4$

$\Longrightarrow (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....... )*\log_{2}(x) = 4$

$\Longrightarrow 2*\log_{2}(x) = 4 \Longrightarrow \log_{2}(x) = 2 \Longrightarrow x = 2^{2} = \boxed{4}$ .

@Sandeep Bhardwaj Nice question, but I think you have omitted a term in the equation; please see my note above.

Following logarithmic laws: Log(x)^k = klog(x) .You are using a square root on each term which. This means that each term goes up to the power of a half from the last one so log2(x) + log2(x)^1/2 ....=4 so using the sum to infinity equation: Sn= a(1-r^n)/ 1-r where a = the first term (log2(x)) and r= 1/2 then it becomes 2= log2(x) and then 2^2 =x which is 4!