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(C) Given equation: $\log _{2}\left(x^{2}+2 x-1\right)=1$By the definition of logarithm,$\log _{b}(a) = c \implies a = b^{c}$.So,$x^{2}+2 x-1 = 2^{1}

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$2$

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(C) Given equation: $\log _{2}\left(x^{2}+2 x-1\right)=1$By the definition of logarithm,$\log _{b}(a) = c \implies a = b^{c}$.So,$x^{2}+2 x-1 = 2^{1} = 2$.Rearranging the terms: $x^{2}+2 x-3 = 0$.Factoring the quadratic equation: $(x+3)(x-1) = 0$.This gives $x = -3$ or $x = 1$.Check the domain: For $\log _{2}(f(x))$ to be defined,$f(x) > 0$.If $x = 1$,$x^{2}+2 x-1 = 1+2-1 = 2 > 0$ (Valid).If $x = -3$,$x^{2}+2 x-1 = 9-6-1 = 2 > 0$ (Valid).Both solutions are valid,so there are $2$ solutions.

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

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