try using $\displaystyle \log_aN=\frac{1}{\log_Na}$ for $\displaystyle \frac{p+q}{p-q}$
EDIT:
forgot to mention
you need $\displaystyle \log_ab=\frac{\log_Nb}{\log_Na}$ as well. That is where "N" disappears.
Hello, MichaelLight!
My approach is worse that Pickslides' suggestion.
I used the Base-Change formula.
$\displaystyle \text{If }\,\log_4 N=p\,\text{ and }\,\log_{12}N=q,\,\text{ show that: }\:\log_3 48\:=\: \dfrac{p+q}{p-q}$
$\displaystyle \displaystyle\text{We have: }\:\log_348 \;\;=\;\;\frac{\ln48}{\ln3} \;\;=\;\;\frac{\ln(12\cdot4)}{\ln(\frac{12}{4})} \;\;=\;\;\frac{\ln12 + \ln4}{\ln12 - \ln4}$
Divide numerator and denominator by $\displaystyle (\ln12\!\cdot\!\ln4)\!:$
. . $\displaystyle \displaystyle =\;\;\frac{\;\dfrac{\ln12 + \ln4}{\ln12\cdot\ln4}\;} {\dfrac{\ln12 - \ln4}{\ln12\cdot\ln4}} \;\;=\;\;\frac{\;\dfrac{1}{\ln4} + \dfrac{1}{\ln12}\;}{\dfrac{1}{\ln4} - \dfrac{1}{\ln12}}$
Multiply by $\displaystyle \dfrac{\ln N}{\ln N}\!:$
. . $\displaystyle \displaystyle =\;\;\frac{\ln N}{\ln N}\cdot\frac{\dfrac{1}{\ln4} + \dfrac{1}{\ln12}}{\dfrac{1}{\ln4} - \dfrac{1}{\ln12}} \;\;=\;\;\frac{\;\dfrac{\ln N}{\ln 4} + \dfrac{\ln N}{\ln12}\;}{\dfrac{\ln N}{\ln 4} - \dfrac{\ln N}{\ln 12}} $
. . $\displaystyle \displaystyle =\;\;\frac{\log_4\!N + \log_{12}\!N}{\log_4\!N - \log_{12}\!N}} \;\;=\;\;\frac{p + q}{p - q} $