Simplifying Logarithmic Expression

$p = \log_{12} 81 = \frac{4\log 3}{\log 12} = \frac{4\log 3}{2\log 2 + \log 3}$. $4 - p = 4 - \frac{4\log 3}{2\log 2 + \log 3} = \frac{8\log 2}{2\log 2 + \log 3}$. $4 + p = 4 + \frac{4\log 3}{2\log 2 + \log 3} = \frac{8\log 2 + 8\log 3}{2\log 2 + \log 3}$. $\frac{4-p}{4+p} = \frac{8\log 2}{8(\log 2 + \log 3)} = \frac{\log 2}{\log 6} = \log_6 2$. $3 \left(\frac{4-p}{4+p}\right) = 3 \log_6 2 = \log_6 8$.