Nested Logarithms

The given expression simplifies to

$\log_{2^{\frac{1}{2}}}{x} + \log_{2^{\frac{1}{4}}}{x} + \cdots + \log_{2^{\frac{1}{128}}}{x} = 1016$ .

We know that $\log_{a}{m}= \dfrac{\log_{b}{m}}{\log_{b}{a}}$ If $a \neq 1$ , $b \neq 1$ . We take the base $b$ as $2$ .

$\implies \dfrac{\log_{2}{x}}{\log_{2}{2^{\frac{1}{2}}}} + \dfrac{\log_{2}{x}}{\log_{2}{2^{\frac{1}{4}}}} + \cdots + \dfrac{\log_{2}{x}}{\log_{2}{2^{\frac{1}{128}}}} = 1016$

$\implies \dfrac{\log_{2}{x}}{\dfrac{1}{2}} + \dfrac{\log_{2}{x}}{\dfrac{1}{4}} + \cdots + \dfrac{\log_{2}{x}}{\dfrac{1}{128}}=1016$ .

$\implies 2\times \log_{2}{x} + 4\times \log_{2}{x} + 8\times \log_{2}{x} + \cdots + 128 \times \log_{2}{x}=1016$ .

$\implies \log_{2}{x}\times \left(2+4+8+16+32+64+128\right) =1016$ .

$\implies \log_{2}{x} \times 254 = 1016$ .

$\implies \log_{2}{x} = 4$ .

$\implies x = 16$

$\begin{aligned} \log_{\sqrt2}{x} + \log_{\sqrt{\sqrt2}}{x} + \log_{\sqrt{\sqrt{\sqrt2}}}{x} +\cdots + \log_{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt 2}}}}}}}x & = 1016 \\ \frac {\log x}{\frac 12 \log 2} + \frac {\log x}{\frac 14 \log 2} + \frac {\log x}{\frac 18 \log 2} + \cdots + \frac {\log x}{\frac 1{2^7} \log 2} & = 1016 \\ \frac {\log x}{\log 2} (2+2^2+2^3+\cdots + 2^7) & = 1016 \\ \frac {\log x}{\log 2} \times 2 \times \frac {2^7-1}{2-1} & = 1016 \\ \frac {\log x}{\log 2} & = 4 \\ \log x & = 4 \log 2 \\ \implies x & = 2^4 = \boxed{16} \end{aligned}$