[tex]\bf log_{12}(x-3)-log_{12}(x+7)=log_{12}(3)\\\\
-----------------------------\\\\
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\qquad thus\\\\
-----------------------------\\\\
log_{12}\left( \cfrac{x-3}{x+7} \right)=log_{12}(3)\impliedby
\begin{array}{llll}
\textit{removing the log from}\\
\textit{both sides}
\end{array}
\\\\\\
\cfrac{x-3}{x+7}=3\implies x-3=3x+21\implies -24=2x\implies -12=x[/tex]
