Answer
The solution set is $\left\{ \frac{3\ln 7+4\ln 5}{\ln 7-2\ln 5} \right\}$ or $-9.64$ .
Work Step by Step
Consider the given expression,
${{7}^{x-3}}={{5}^{2x+4}}$
Taking the logarithm on both sides, we get,
$\begin{align}
& \log \left( {{7}^{x-3}} \right)=\log \left( {{5}^{2x+4}} \right) \\
& \left( x-3 \right)\ln 7=\left( 2x+4 \right)\ln 5
\end{align}$
It can be further simplified as,
$\begin{align}
& x\ln 7-3\ln 7=2x\ln 5+4\ln 5 \\
& x\ln 7-2x\ln 5=4\ln 5+3\ln 7 \\
& x\left( \ln 7-2\ln 5 \right)=4\ln 5+3ln7 \\
& x=\frac{4\ln 5+3ln7}{\ln 7-2\ln 5}
\end{align}$
This can be approximated as:
$\begin{align}
& x=\frac{4\ln 5+3ln7}{\ln 7-2\ln 5} \\
& =\frac{6.43775165+5.83773045}{1.945910-3.218876} \\
& =-9.64
\end{align}$
