Interesting logarithms - 1

$\begin{cases} \log_m w = \dfrac {\log w}{\log m} = 12 & \implies \log m = \dfrac {\log w}{12} \\ \log_n w = \dfrac {\log w}{\log n} = 16 & \implies \log n = \dfrac {\log w}{16} \\ \log_p w = \dfrac {\log w}{\log p} = 36 & \implies \log p = \dfrac {\log w}{36} \end{cases}$

Then

$\begin{aligned} \log_{mnpq} w & = 72 \\ \frac {\log w}{\log (mnpq)} & = 72 \\ 72 \log (mnpq) & = \log w \\ 72(\log m + \log n + \log p + \log q) & = \log w \\ 72 \left(\frac 1{12} + \frac 1{16} + \frac 1{36} \right)\log w + 72 \log q & = \log w \\ \left(6 + \frac 92 + 2 \right)\log w + 72 \log q & = \log w \\ 72 \log q & = \log w \left(1 - \frac {25}2\right) = - \frac {23}2 \log w \\ \frac {\log w}{\log q} & = - \frac {144}{23} \\ \implies - \log_q w & = \frac {144}{23} \end{aligned}$

Therefore, $a^2 - b^2 = 144^2 - 23^2 = \boxed{20207}$ .

I will be using these equations again and again through out the solution : $\log_m w=12$ $\log_n w=16$ $\log_p w=36$ $\log_{mnpq}w=72$

From our $4_{th}$ equation : $m^{72}n^{72}p^{72}q^{72}=w$ $(m^{12})^{6}(n^{16})^{\frac{9}{2}}(p^{36})^{2}q^{72}=w$ Simplifying using our $1_{st}, 2_{nd}$ and $3_{rd}$ equation $w^6w^{\frac{9}{2}}w^2q^{72}=w$ $q^{72}w^{2+6+\frac{9}{2}}=w$ $q^{72}w^{\frac{25}{2}}=w$ $q^{72}=w^{1-\frac{25}{2}}$ $q^{72}=w^{\frac{-23}{2}}$ Squaring both sides $q^{144}=w^{-23}$ Raising both sides to the power of $\frac{-1}{23}$ $q^{\frac{-144}{23}}=w$ Therefore, $log_qw=\frac{-144}{23}$ $-\log_q w=\frac{144}{23}$ As $144$ and $23$ are co-prime therefore evaluating for the expression we wanted : $144^2-23^2=(144-23)(144+23)=(121)(167)=\boxed{20207}$