Question

The number of solutions of the equation ${\log }_2\left(x^2+2x-1\right)=1$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Given ${\log }_2\left(x^2+2x-1\right)=1$
$\Rightarrow {\log }_2\left(x^2+2x-1\right)={\log }_{2}2$
$\Rightarrow x^2+2x-1=2$
$\Rightarrow x^2+2x-3=0$
$\Rightarrow x^2+3x-x-3=0$
$\Rightarrow x(x+3)-1(x+3)=0$
$\Rightarrow (x+3)(x-1)=0$
$\Rightarrow x=1,-3$
Since, $x=1$ and $x=-3$ are satisfy the given equation therefore the number of solutions of the equation are two.