The number of solution(s) of the equation ${log}_{2} (x^{2} + 2 x - 1) = 1$ is/are

0

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2

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3

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Solution

The correct option is C $2$
$x^{2} + 2 x - 1 > 0$ ------------------------(1)

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

$\Rightarrow {log}_{2} (x^{2} + 2 x - 1) = {log}_{2} 2$
$\Rightarrow x^{2} + 2 x - 1 = 2$
$\Rightarrow x^{2} + 2 x - 3 = 0$
$\Rightarrow x^{2} + 3 x - x - 3 = 0$

$\Rightarrow x (x + 3) - 1 (x + 3) = 0$
$\Rightarrow (x + 3) (x - 1) = 0$

$\Rightarrow x = 1, - 3$

Since, $x = 1$ and $x = - 3$ satisfy the given equation and the equation (1). Therefore the number of solutions of the equation are two.

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