Question

Solve the following equation:
$\log (x^3+27)-0.5\log (x^2+6x+9)=3\log \sqrt[3]{37}$

Answer 1

$\log (x^3+27)-\left(0.5\right)\log (x^2+6x+9)=3\log \sqrt[3]{37}\Rightarrow \log (x^3+27)-\left(0.5\right)\log {(x+3)}^2=\log 7\Rightarrow \log \left(\frac{{(x}^3+27)}{x+3}\right)=\log 7\Rightarrow x\ne -3$

Therefore, $\log (x^2+3x+9)=\log 7\Rightarrow x^2-3x+9=7\Rightarrow x^2-3x+2=0$

Therefore, $x=1,2.$