Lösung 3.3:3f
Aus Online Mathematik Brückenkurs 1
If we write \displaystyle \text{4} and \displaystyle \text{16} as
\displaystyle \begin{align}
& \text{4}=2\centerdot 2=2^{2} \\
& 16=2\centerdot 8=2\centerdot 2\centerdot 4=2\centerdot 2\centerdot 2\centerdot 2=2^{4} \\
\end{align}
we obtain
\displaystyle \begin{align}
& \log _{2}4+\log _{2}\frac{1}{16}=\log _{2}2^{2}+\log _{2}\frac{1}{2^{4}} \\
& =\log _{2}2^{2}+\log _{2}2^{-4}=2\centerdot \log _{2}2+\left( -4 \right)\centerdot \log _{2}2 \\
& =2\centerdot 1+\left( -4 \right)\centerdot 1=-2 \\
\end{align}