Proof of a logarithmic equation If \begin{align}\log_{16}{15} &= a\\
\log_{12}{18} &= b\\
\log_{25}{24} &= c\end{align} 
then prove that $$c=\frac{5-b}{2(8a - 4ab -2b +1)}$$
My attempt: I tried to prove it by applying the standard logarithmic formulas such as $$\log_{a}{b} = \log_{c}{b}\times \log_{a}{c} $$ but the denominator is becoming too complex to perform an LCM of all the terms. Can there be a simpler way to prove this?
Any help is appreciated.
 A: Let $\log_52=x$ and $\log_53=y$.
Hence, $$a=\frac{\log_515}{\log_516}=\frac{1+y}{4x},$$
which gives $y=4ax-1$ and
$$b=\frac{\log_518}{\log_512}=\frac{x+2y}{2x+y},$$ which gives
$$(2b-1)x=(2-b)y$$ or
$$y=\frac{(2b-1)x}{2-b}$$ and we obtain
$$x\left(4a-\frac{2b-1}{2-b}\right)=1$$ or
$$x=\frac{2-b}{8a-4ab-2b+1},$$ which gives
$$y=\frac{2b-1}{8a-4ab-2b+1}.$$
Id est,
$$c=\frac{1}{2}(3x+y)=\frac{3(2-b)+2b-1}{2(8a-4ab-2b+1)}=\frac{5-b}{2(8a-4ab-2b+1)}.$$
A: Without using logs, purely by manipulation of indices:
$$\begin{align}
12^b&=18\\
2^{2b}\cdot 3^b&=2\cdot 3^2\\
2^{2b-1}&=3^{2-b}\\
3&=2^{\frac{2b-1}{2-b}}\tag{1}\\
\text{Also,}\hspace{3cm}\\
16^a&=15\\
2^{4a}&=3\cdot 5=2^{\frac{2b-1}{2-b}}\cdot 5
&&\scriptsize(\text{using}\;\; (1))\\
2^{4a-\frac {2b-1}{2-b}}&=5\\
2^{\frac {8a-4ab-2b+1}{2-b}}&=5\\
2&=5^\frac{2-b}{8a-4ab-2b+1}\tag{2}\\
\text{And finally, }\hspace{3cm}\\
25^c&=24\\
5^{2c}&=3\cdot 2^3\\
&=2^{\frac{2b-1}{2-b}}\cdot 2^3
&&\scriptsize(\text{using}\;\; (1))\\
&=2^{\frac {2b-1}{2-b}+3}\\
&=2^\frac{5-b}{2-b}\\
&=\left(5^\frac{2-b}{8a-4ab-2b+1}\right)^\frac{5-b}{2-b}
&&\scriptsize(\text{using}\;\; (2))\\\\
&=5^\frac{5-b}{8a-4ab-2b+1}\\
\text{Equating indices and dividing by $2$,}\\
\color{red}c&\color{red}{=\frac{5-b}{2(8a-4ab-2b+1)}\;\;\blacksquare}
\end{align}$$
