If 2 log (x+1)-log (x^{2}-1)=log 2,then x= | vedclass

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(A) Given the equation: $2 \log (x+1)-\log (x^{2}-1)=\log 2$. Using the property $n \log a = \log (a^n)$,we get $\log (x+1)^2 - \log (x^2-1) = \log 2$

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only $3$

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(A) Given the equation: $2 \log (x+1)-\log (x^{2}-1)=\log 2$. Using the property $n \log a = \log (a^n)$,we get $\log (x+1)^2 - \log (x^2-1) = \log 2$. Using the property $\log a - \log b = \log (a/b)$,we get $\log \left( \frac{(x+1)^2}{x^2-1} \right) = \log 2$. Since $x^2-1 = (x-1)(x+1)$,the equation becomes $\log \left( \frac{(x+1)^2}{(x-1)(x+1)} \right) = \log 2$. Simplifying the fraction,we have $\log \left( \frac{x+1}{x-1} \right) = \log 2$. Therefore,$\frac{x+1}{x-1} = 2$. $x+1 = 2(x-1) \Rightarrow x+1 = 2x-2$. $x = 3$. For the logarithmic terms to be defined,we require $x+1 > 0$ and $x^2-1 > 0$,which implies $x > 1$. Thus,$x = 3$ is the only valid solution.

If $2 \log (x+1)-\log (x^{2}-1)=\log 2$,then $x=$

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