Solve the equation: $\;$ $\log \left(3x^2 + 7\right) - \log \left(3x - 2\right) = 1$
Given equation: $\;\;$ $\log \left(3x^2 + 7\right) - \log \left(3x - 2\right) = 1$
i.e. $\;$ $\log \left(\dfrac{3x^2 + 7}{3x - 2}\right) = 1$
i.e. $\;$ $\dfrac{3x^2 + 7}{3x - 2} = 10^1 = 10$
i.e. $\;$ $3x^2 + 7 = 30x - 20$
i.e. $\;$ $3x^2 - 30x + 27 = 0$
i.e. $\;$ $x^2 - 10x + 9 = 0$
i.e. $\;$ $\left(x - 9\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = 9$, $\;$ or $\;$ $x = 1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1, 9 \right\}$