Question

Solve the following equation:
$\frac{1}{3}\log (x^2-16x+20)-\log \sqrt[3]{37}=\frac{1}{3}\log (8-x)$

Answer 1

$\frac{1}{3}log(x^2-16+20)-log\sqrt[3]{37}=\frac{1}{3}log(8-x)log(x^2-16x+20)-3log\sqrt[3]{37}=log(8-x)\frac{x^2-16x+20}{7}=8-x{x}^2-16x+20=56-7x{x}^2-9x-36=0$

$x=-3,12$

For $x=12,8-x=-4<0$

But $8-x$ must be $>0$

$\therefore x=-3$ is only possible solution.