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

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(C) If $a, b, c$ are in $A.P.,$ then $2b = a + c.$$\Rightarrow 2 \log _{10}(2^x - 1) = \log _{10} 2 + \log _{10}(2^x + 3)$$\Rightarrow \log _{10}(2^x

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$x = \log _{2} 5$

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(C) If $a, b, c$ are in $A.P.,$ then $2b = a + c.$$\Rightarrow 2 \log _{10}(2^x - 1) = \log _{10} 2 + \log _{10}(2^x + 3)$$\Rightarrow \log _{10}(2^x - 1)^2 = \log _{10} [2(2^x + 3)]$$\Rightarrow (2^x - 1)^2 = 2(2^x + 3)$Let $2^x = y.$$\Rightarrow (y - 1)^2 = 2(y + 3)$$\Rightarrow y^2 - 2y + 1 = 2y + 6$$\Rightarrow y^2 - 4y - 5 = 0$$\Rightarrow (y - 5)(y + 1) = 0$Since $y = 2^x > 0,$ we have $y = 5.$$\Rightarrow 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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