An algebra problem by Saksham Jain

$=\left( \huge \log_{128}2 \right)^{ \log_{\frac{1}{2^{3}}+\frac{1} {2^6}+\frac{1}{2^9}+\ldots}4 } \\ = \left(\huge \frac{1}{7}\log_{2}2\right)^{\log_{\left(\frac{\frac{1}{2^3}}{1-\frac{1}{2^3}}\right)}4} \\ \text{We know, } a^{\log_{a}b}=b \\ \therefore \left( \huge \frac{1}{7} \right)^{\log_{\frac{1}{7}}4} = \boxed{4}$

$A=\sum_{n=1}^{\infty}\frac{1}{2^{3n}}=\sum_{n=1}^{\infty}\frac{1}{8^n}=\sum_{n=0}^{\infty}\left ( \frac{1}{8} \right )^n-1.$

$Since\; \left | \frac{1}{8} \right |< 1, \; A\;converges\;to$

$\frac{1}{1-\frac{1}{8}}-1=\frac{1}{7}$

If $B=log_{128}({2})=\frac{log_{2}{(2)}}{log_{2}{(128)}}=\frac{1}{7}=A$

then $B^{log_{A}{4}}=B^{log_{B}{4}}=\boxed{4}$