# Solution to Practice Problem #7

\begin{array}{c} \eqalign{ & \left( {{{\log }_a}b} \right)\left( {{{\log }_b}a} \right) = 1 \cr & \cr & = \left( {{\text{lo}}{{\text{g}}_a}b} \right)({\log _b}a) \cr & = \frac{{\ln b}}{{\ln a}} \cdot \frac{{\ln a}}{{\ln b}} \cr & = 1 \cr} \end{array}

\begin{array}{c} \eqalign{ & {\log _{{a^n}}}b = \frac{{{{\log }_a}b}}{n} \cr & \cr & = {\log _{{a^n}}}b \cr & = \frac{{{{\log }_a}b}}{{{{\log }_a}{a^n}}} \cr & = \frac{{{{\log }_a}b}}{{n\,{{\log }_a}a}} \cr & = \frac{{{{\log }_a}b}}{n} \cr} \end{array}

\begin{array}{c} \eqalign{ & {\log _b}\left( {{a^{{{\log }_b}c}}} \right) = {\log _b}\left( {{c^{{{\log }_b}a}}} \right) \cr & {\log _b}c \cdot {\log _b}a = {\log _b}a \cdot {\log _b}c \cr & \therefore \,\,{\text{true}} \cr} \end{array}

\begin{array}{c} \eqalign{ & {\log _a}\frac{1}{b} = {\log _{\frac{1}{a}}}b \cr & \cr & \frac{{\ln \frac{1}{b}}}{{\ln a}} = \frac{{\ln b}}{{\ln \frac{1}{a}}} \cr & \frac{{ – \ln b}}{{\ln a}} = \frac{{\ln b}}{{ – \ln a}} \cr & \therefore \,\,{\text{true}} \cr} \end{array}