Given:
[tex]\log _97=a,\log _93=b[/tex]
The change of base formula is,
[tex]\log _b(a)=\frac{\log_xa}{\log_xb}[/tex]
a)
[tex]\begin{gathered} \log _39 \\ \text{Use:}\log _bc=\frac{1}{\log_cb} \\ \log _39=\frac{1}{\log_93} \\ \text{Given: }\log _93=b \\ \log _39=\frac{1}{b} \end{gathered}[/tex]
b)
[tex]\begin{gathered} \log _3(\frac{7}{3}) \\ \text{Use:}\log _a\frac{l}{m}=\log _al-\log _am \\ \log _3(\frac{7}{3})=\log _37-\log _33 \\ We\text{ know, }\log _aa=1 \\ \log _3(\frac{7}{3})=\log _37-1 \\ \log _3(\frac{7}{3})=\frac{\log_97}{\log_93}-1\ldots\ldots...\text{ Change of base property} \\ \log _3(\frac{7}{3})=\frac{a}{b}-1 \end{gathered}[/tex]
Answer:
[tex]\begin{gathered} \log _39=\frac{1}{b} \\ \log _3(\frac{7}{3})=\frac{a}{b}-1 \end{gathered}[/tex]
