Question

Solve the following equation:
$\frac{1}{2}\log x+3\log \sqrt{2+x}=\log \sqrt{x(x+2)}+2$

Answer 1

$\frac{1}{2}\log x+3\log \sqrt{2+x}=\log \sqrt{x(x+2)}+2\Rightarrow \log \left[\right(x+2\left)\sqrt{x(x+2)}\right]=\log \left[100\sqrt{x(x+2)}\right]\Rightarrow (x+2)\sqrt{x(x+2)}=100\sqrt{x(x+2)}\Rightarrow \sqrt{x(x+2)}=0$

or $x+2=100$

Therefore, $x=0,-2$

or $x=98$

But $x$ cannot be $\le 0$

Therefore, $\log x$ is undefined fr $x\le 0$

Therefore, $x=98$ is only solution.