<s>[INST] 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.
 [/INST] 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,
$$c=\frac{1}{2}(3x+y)=\frac{3(2-b)+2b-1}{2(8a-4ab-2b+1)}=\frac{5-b}{2(8a-4ab-2b+1)}.$$
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