what's the value of log2 (1/8) ?
log2(1/8)=-log2(8)=-(log2(2))^3=-3log2(2)-3
\(log_aa^p=p\\------\\\\log_2 ( \frac{1}{8} ) =log_28^{-1}=log_2(2^3)^{-1}=log_22^{-3}=-3\\\\Ans.\ -3\)
