$\begin{array}{l} c=\log_{b} a \Leftrightarrow b^{c}=a \\ \log_c a+\log_c b = \log_c (a \cdot b) \\ \log_c a-\log_c b =\log _c\frac{a}{b} \\ log_c a^n=n\log_c a \\ \\ \textbf{Gegeben:} \\ {a=3\frac{11}{25} \qquad b=3\frac{2}{5} \qquad c=4 \qquad n=5}\\ \\ \textbf{Rechnung:} \\ \log_{3\frac{2}{5}} 3\frac{11}{25} =1,01 \Leftrightarrow 3\frac{2}{5}^{1,01}=3\frac{11}{25} \\ \log_{4} 3\frac{11}{25}+\log_{4}3\frac{2}{5} = \log_{4}(3\frac{11}{25} \cdot 3\frac{2}{5})= \log_{4}(3\frac{11}{25} \cdot 3\frac{2}{5})=1,77 \\ \log_{4} 3\frac{11}{25}-\log_{4}3\frac{2}{5} =\log_{4}\frac{3\frac{11}{25}}{3\frac{2}{5}}= 0,00844\\ \log_{4}3\frac{11}{25}^5=5\log_{4}3\frac{11}{25} = 4,46\\ \log_{3\frac{2}{5}} 3\frac{11}{25}=\dfrac{\log_{4}3\frac{11}{25}}{\log_{4}3\frac{2}{5}}=1,01 \end{array}$