Algebra-Grundlagen-Logarithmen

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Beispiel Nr: 02
$\begin{array}{l} c=\log_{b} a \Leftrightarrow b^{c}=a \\ \log_c a+\log_c b = \log_c (a \cdot b) \\ \log_c a-\log_c b =\log _c\frac{a}{b} \\ log_c a^n=n\log_c a \\ \\ \textbf{Gegeben:} \\ {a=2 \qquad b=3 \qquad c=2 \qquad n=4}\\ \\ \textbf{Rechnung:} \\ \log_{3} 2 =0,631 \Leftrightarrow 3^{0,631}=2 \\ \log_{2} 2+\log_{2}3 = \log_{2}(2 \cdot 3)= \log_{2}(2 \cdot 3)=2,58 \\ \log_{2} 2-\log_{2}3 =\log_{2}\frac{2}{3}= -0,585\\ \log_{2}2^4=4\log_{2}2 = 4\\ \log_{3} 2=\dfrac{\log_{2}2}{\log_{2}3}=0,631 \end{array}$