The logarithm of $.0625$ to the base $2$ is:
$ \textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2 $
1. We start with the given logarithmic expression: 
   \[
   \log_2 0.0625
   \]
2. Convert \(0.0625\) to a fraction:
   \[
   0.0625 = \frac{1}{16}
   \]
3. Rewrite the logarithmic expression using the fraction:
   \[
   \log_2 \left(\frac{1}{16}\right)
   \]
4. Recall that \(\frac{1}{16}\) can be written as \(2^{-4}\) because:
   \[
   16 = 2^4 \implies \frac{1}{16} = 2^{-4}
   \]
5. Substitute \(2^{-4}\) into the logarithmic expression:
   \[
   \log_2 \left(2^{-4}\right)
   \]
6. Use the property of logarithms that \(\log_b (b^x) = x\):
   \[
   \log_2 \left(2^{-4}\right) = -4
   \]

The final answer is \(\boxed{-4}\).