$I f \log_{5}^{2}, \log_{5} (2^{x} - 5) a n d \log_{5} (2^{x} - \frac{7}{2}) a r e i n A . P . T h e n x =$

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4 or 5

1/2 or 3/2

answer is B.

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Detailed Solution

$\log_{5}^{2}, \log_{5} (2^{x} - 5) a n d \log_{5} (2^{x} - \frac{7}{2}) a r e i n A P \Rightarrow 2 \log_{5} (2^{x} - 5) = \log_{5}^{2} + \log_{2} (2^{x} - \frac{7}{2})$

$\Rightarrow {(2^{x} - 5)}^{2} = 2 (2^{x} - \frac{7}{2}) \Rightarrow {(2^{x})}^{2} - 12 (2^{x}) + 32 = 0$

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