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If $\log_{10} 2, \log_{10} (2^{x} + 1)$ and $\log_{10} (2^{x} + 3)$ are in A.P., then

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x = 0

x = 1

x=12log10⁡2

x=12log2⁡5

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detailed solution

Correct option is D

2,2x+1,2x+3 are in G.P.⇒2x+12=22x+3⇒22x=5⇒x=12log2⁡5

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If log10⁡2,log10⁡2x+1 and log10⁡2x+3 are in A.P., then

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If $\log_{10} 2, \log_{10} (2^{x} + 1)$ and $\log_{10} (2^{x} + 3)$ are in A.P., then