Lösung 3.3:3f

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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}