If log _{10} 2, log _{10} (2^x - 1), log _{10 | vedclass

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(D) If $a, b, c$ are in $A.P.$,then $2b = a + c$.Given terms are $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$.Applying the condition: $2

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$x = \log _{2} 5$

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(D) If $a, b, c$ are in $A.P.$,then $2b = a + c$.Given terms are $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$.Applying the condition: $2 \log _{10} (2^x - 1) = \log _{10} 2 + \log _{10} (2^x + 3)$.Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:$\log _{10} (2^x - 1)^2 = \log _{10} [2(2^x + 3)]$.Removing the logarithm from both sides:$(2^x - 1)^2 = 2(2^x + 3)$.Let $2^x = y$. Then $(y - 1)^2 = 2(y + 3)$.$y^2 - 2y + 1 = 2y + 6$.$y^2 - 4y - 5 = 0$.$(y - 5)(y + 1) = 0$.So,$y = 5$ or $y = -1$.Since $y = 2^x$ must be positive,$y = 5$.$2^x = 5 \Rightarrow x = \log _{2} 5$.

If $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

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