Question

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

• A. $-2+\sqrt{3}$
• B. $-4+2\sqrt{3}$
• C. $-2-\sqrt{3}$
• D. $-4-2\sqrt{3}$

Answer 1

Answer: (A), (C)

$-2+\sqrt{3}$
$-2-\sqrt{3}$

We have,

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

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

Comparing both side and we get,

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

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

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

Using quadratic formula and we get,

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

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

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

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

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

Hence, this is the answer.