Lösung 3.3:3f

4 &= 2\cdot 2 = 2^2\,,\\[5pt] 16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,, \end{align}

\log_2 4 + \log_2\frac{1}{16} &= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt] &= \log_2 2^2 + \log_2 2^{-4}\\[5pt] &= 2\cdot\log_2 2 + (-4)\cdot\log_2 2\\[5pt] &= 2\cdot 1 + (-4)\cdot 1\\[5pt] &= -2\,\textrm{.} \end{align}