Algebra-Grundlagen-Logarithmen



Beispiel Nr: 04
$c=\log_{b} a \Leftrightarrow b^{c}=a \\ \log_c a+\log_c b = \log_c (a \cdot b) \\ \log_c a-\log_c b =\log _c\frac{a}{b} \\ log_c a^n=n\log_c a \\ \\ \textbf{Gegeben:} \\ {a=2\frac{1}{2} \qquad b=4\frac{1}{8} \qquad c=2 \qquad n=2}\\ \\ \textbf{Rechnung:} \\ \log_{4\frac{1}{8}} 2\frac{1}{2} =0,647 \Leftrightarrow 4\frac{1}{8}^{0,647}=2\frac{1}{2} \\ \log_{2} 2\frac{1}{2}+\log_{2}4\frac{1}{8} = \log_{2}(2\frac{1}{2} \cdot 4\frac{1}{8})= \log_{2}(2\frac{1}{2} \cdot 4\frac{1}{8})=3,37 \\ \log_{2} 2\frac{1}{2}-\log_{2}4\frac{1}{8} =\log_{2}\frac{2\frac{1}{2}}{4\frac{1}{8}}= -0,722\\ \log_{2}2\frac{1}{2}^2=2\log_{2}2\frac{1}{2} = 2,64\\ \log_{4\frac{1}{8}} 2\frac{1}{2}=\dfrac{\log_{2}2\frac{1}{2}}{\log_{2}4\frac{1}{8}}=0,647 $