Solve the equation: $\;$ $\dfrac{1}{2} \log x + 3 \log \sqrt{2 + x} = \log \sqrt{x \left(x + 2\right)} + 2$
Given equation: $\;\;$ $\dfrac{1}{2} \log x + 3 \log \sqrt{2 + x} = \log \sqrt{x \left(x + 2\right)} + 2$
i.e. $\;$ $\log \sqrt{x} - \log \sqrt{x \left(x + 2\right)} + \log \left(\sqrt{2 + x}\right)^3 = 2$
i.e. $\;$ $\log \left[\dfrac{\sqrt{x} \left(2 + x\right) \sqrt{2 + x}}{\sqrt{x \left(x + 2\right)}}\right] = 2$
i.e. $\;$ $\log \left(x + 2\right) = 2$
i.e. $\;$ $x +2 = 10^2 = 100$
i.e. $\;$ $x = 98$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{98 \right\}$