If log 2,log (2^x - 1),and log (2^x + 3) are | vedclass

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(B) Given that $\log 2$,$\log (2^x - 1)$,and $\log (2^x + 3)$ are in arithmetic progression $(AP)$.By the property of $AP$,$2 	imes 	ext{middle term

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$\log_2 5$

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(B) Given that $\log 2$,$\log (2^x - 1)$,and $\log (2^x + 3)$ are in arithmetic progression $(AP)$.By the property of $AP$,$2 	imes 	ext{middle term} = 	ext{first term} + 	ext{third term}$.$2 \log (2^x - 1) = \log 2 + \log (2^x + 3)$Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:$\log (2^x - 1)^2 = \log [2(2^x + 3)]$$(2^x - 1)^2 = 2(2^x + 3)$Let $2^x = y$. Then $(y - 1)^2 = 2(y + 3)$$y^2 - 2y + 1 = 2y + 6$$y^2 - 4y - 5 = 0$$(y - 5)(y + 1) = 0$Since $y = 2^x > 0$,we have $y = 5$.$2^x = 5 \Rightarrow x = \log_2 5$.

If $\log 2$,$\log (2^x - 1)$,and $\log (2^x + 3)$ are in arithmetic progression,then the value of $x$ is:

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