If \log\ (-2x)=2\log\ (x+1), then $x$ can be  equal to

We have,

$\log \left(-2x\right)=2\log \left(x+1\right)$

$\Rightarrow \log \left(-2x\right)=\log {\left(x+1\right)}^2$

Comparing both side and we get,

$-2x={\left(x+1\right)}^2$

$\Rightarrow -2x=x^2+1+2x$

$\Rightarrow x^2+4x+1=0$

Using quadratic formula and we get,

$x=\frac{-4\pm \sqrt{16-4\times 1\times 1}}{2\times 1}$

$x=\frac{-4\pm \sqrt{12}}{2}$

$x=\frac{-4\pm \sqrt{2\times 2\times 3}}{2}$

$x=\frac{-4\pm 2\sqrt{3}}{2}$

$x=-2\pm \sqrt{3}$