If log 2, log (2^{x}-1) and log (2^{x}+3) (al | vedclass

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(C) Given that $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ are in an Arithmetic Progression $(A.P.)$.For three terms $a, b, c$ to be in $A.P.$,the c

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$\log _{2} 5$

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(C) Given that $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ are in an Arithmetic Progression $(A.P.)$.For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.Therefore,$2 \log (2^{x}-1) = \log 2 + \log (2^{x}+3)$.Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^{n})$,we get:$\log (2^{x}-1)^{2} = \log [2(2^{x}+3)]$.Removing the logarithms from both sides:$(2^{x}-1)^{2} = 2(2^{x}+3)$.Let $2^{x} = y$. Then the equation becomes:$(y-1)^{2} = 2(y+3)$.$y^{2} - 2y + 1 = 2y + 6$.$y^{2} - 4y - 5 = 0$.Factoring the quadratic equation:$(y-5)(y+1) = 0$.So,$y = 5$ or $y = -1$.Since $2^{x} = y$ and $2^{x}$ must be positive for all real $x$,we discard $y = -1$.Thus,$2^{x} = 5$.Taking the logarithm on both sides with base $2$:$x = \log _{2} 5$.

If $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ (all to the base $10$) are three consecutive terms of an Arithmetic Progression,then the value of $x$ is equal to

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