Logarithmic problem If $\log_{16} 15 =a$ and $\log_{12} 18 =b$, then show that $$\log_{25} 24 = \frac{5-b}{16a-8ab-4b+2}.$$

Let $l_2= \ln 2$,$l_3= \ln 3$ and $l_5= \ln 5$. Now change base & we have
\begin{eqnarray*}
\frac{5-b}{16a-8ab-4b+2} = \frac{5-\frac{l_2+2l_3}{2l_2+l_3}}{16 \frac{l_3+l_5}{4l_2}-8 \frac{l_3+l_5}{4l_2} \frac{l_2+2l_3}{2l_2+l_3} -4 \frac{l_2+2l_3}{2l_2+l_3} +2} \\
\end{eqnarray*}
\begin{eqnarray*}
=  \frac{(5(2l_2+l_3)-(l_2+2l_3))l_2} {4 (l_3+l_5)(2l_2+l_3) -2 (l_3+l_5)(l_2+2l_3) -4 (l_2+2l_3)l_2 +2(2l_2+l_3)l_2} \\
\end{eqnarray*}
\begin{eqnarray*}
=  \frac{(9l_2+3l_3)l_2} {(l_3+l_5)6l_2-6l_2l_3} \\
\end{eqnarray*}
\begin{eqnarray*}
=  \frac{(3l_2+l_3)} {2 l_5} =  \log_{25}(24) \\
\end{eqnarray*}
