Question

Solve the following equation: log2(16x)= 4

Answer 1

`\begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{ \textit{we'll use this one} }{a^(log_a (x))=x} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \log_2(16x)=4\implies 2^(\log_2(16x))=2^4\implies 16x=2^4\implies x=\cfrac{2^4}{16}\implies x=1`