Question

Solve $\log (-x)=2\log (x+1)$.
How many solutions of x are possible?

Answer 1

By defination, $x<0$ and $x+1>0\Rightarrow -1<x<0$
Now $\log (-x)=2\log (x+1)$
or $-x={(x+1)}^2$
or $x^2+3x+1=0$
or $x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$
Hence, $x=\frac{-3+\sqrt{5}}{2}$ is the only solution
Ans: 1