How should I go about solving this? Is it even possible?
32/48 = (6^b)/(14^b)
I hope you know logarithms and the rules of logs
$\displaystyle \frac{32}{48}=\frac{6^b}{14^b}$ so
$\displaystyle \log\frac{32}{48}=\log\frac{6^b}{14^b}$ so
$\displaystyle \log\frac{32}{48}=\log 6^b-\log 14^b$
$\displaystyle \log\frac{32}{48}=b\log 6-b \log 14=b(\log 6-\log 14)$
$\displaystyle \log\frac{32}{48}=b(\log\frac{6}{14})$
so yeah... gotta use logarithms