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

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(B) Given that $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$By the property of $A.P.$,if $a, b, c$ are in $A.P.$,then $2b = a + c$.There

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$\log_2 5$

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(B) Given that $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$By the property of $A.P.$,if $a, b, c$ are in $A.P.$,then $2b = a + c$.Therefore,$2 \log (2^n - 1) = \log 2 + \log (2^n + 3)$.Using the logarithmic property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$,we get:$\log (2^n - 1)^2 = \log [2(2^n + 3)]$.Removing the logarithms from both sides:$(2^n - 1)^2 = 2(2^n + 3)$.Let $x = 2^n$. Then $(x - 1)^2 = 2(x + 3)$.$x^2 - 2x + 1 = 2x + 6$.$x^2 - 4x - 5 = 0$.Factoring the quadratic equation: $(x - 5)(x + 1) = 0$.So,$x = 5$ or $x = -1$.Since $x = 2^n$ must be positive,we have $2^n = 5$.Taking the logarithm on both sides,$n = \log_2 5$.

If $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$,then $n =$

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