![\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log(6+7x)-\log(3x-2)=\log(5)\implies \log\left( \cfrac{6+7x}{3x-2} \right)=\log(5) \\\\\\ \cfrac{6+7x}{3x-2}=5\implies 6+7x=15x-10\implies 6=8x-10 \\\\\\ 16=8x\implies \cfrac{16}{8}=x\implies 2=x](/tpl/images/2626/1094/6f8d8.png)
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