Notice that:
[tex]\begin{gathered} (\frac{1}{2})^{2x}=(2^{-1})^{2x}=2^{-2x}, \\ 32=2^5\text{.} \end{gathered}[/tex]
Therefore:
[tex]2^{-2x}=2^5\text{.}[/tex]
Now, applying the logarithm base 2 to the above equation we get:
[tex]\begin{gathered} \log _2(2^{-2x})=\log _22^5\text{.} \\ \end{gathered}[/tex]
Now, recall that:
[tex]\log _bb^m=m\text{.}[/tex]
Therefore:
[tex]-2x=5.[/tex]
Answer: Last option.
[tex]-2x=5.[/tex]
