Logarithm with algebra

$\begin{aligned} \log_{12} 18 & = a \\ \frac {\log 18}{\log 12} & = a \\ \frac {\log 2 + 2\log 3}{2\log 2+\log 3} & = a \\ \frac {1+2 \blue{\frac {\log 3}{\log 2}}}{2+\blue{\frac {\log 3}{\log 2}}} & = a & \small \blue{\text{Let }b = \frac {\log 3}{\log 2}} \\ \frac {1+2 \blue b}{2+\blue b} & = a \\ 1 + 2b & = 2a+ab \\ 2b-ab & 2a-1 \\ \implies b & = \frac {2a-1}{2-a} \end{aligned}$

Now we have:

$\begin{aligned} \log_{24} 16 & = \frac {\log 16}{\log 24} = \frac {4\log 2}{3\log 2+\log 3} = \frac 4{3+\frac {\log 3}{\log 2}} \\ & = \frac 4{3+b} = \frac 4{3+\frac {2a-1}{2-a}} = \frac {8-4a}{6-3a+2a-1} \\ & = \boxed{\frac {8-4a}{5-a}} \end{aligned}$