Solve the equation: $\;$ $\log \left(x^3 + 27\right) - 0.5 \log \left(x^2 + 6x + 9\right) = 3 \log \sqrt[3]{7}$
Given equation: $\;\;$ $\log \left(x^3 + 27\right) - 0.5 \log \left(x^2 + 6x + 9\right) = 3 \log \sqrt[3]{7}$
i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 -3x + 9\right)\right] - 0.5 \log \left(x + 3\right)^2 = \log \left(\sqrt[3]{7}\right)^3$
i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 - 3x + 9\right)\right] - 0.5 \times 2 \log \left(x + 3\right) = \log 7$
i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 - 3x + 9\right)\right] - \log \left(x + 3\right) = \log 7$
i.e. $\;$ $\log \left[\dfrac{\left(x + 3\right) \left(x^2 - 3x + 9\right)}{x + 3}\right] = \log 7$
i.e. $\;$ $\log \left(x^2 - 3x + 9\right) = \log 7$ $\;\;\; \cdots \; (1)$
Taking antilog to base $10$ on both sides of equation $(1)$ gives
$x^2 - 3x + 9 = 7$
i.e. $\;$ $x^2 - 3x + 2 = 0$
i.e. $\;$ $\left(x - 2\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = 2$ $\;$ or $\;$ $x = 1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1, \; 2 \right\}$