If log5⁡2,log5⁡2x−5 and log5⁡2x−7/2 are in A.P., then x is equal to

Solution:

As $\log_{5} 2, \log_{5} (2^{x} - 5), \log_{5} (2^{x} - 7 / 2)$ are in A.P., we get
$\begin{array}{l} 2 \log_{5} (2^{x} - 5) = \log_{5} 2 + \log_{5} (2^{x} - 7 / 2) \\ \Rightarrow {(2^{x} - 5)}^{2} = 2 (2^{x} - 7 / 2) \\ \Rightarrow {(2^{x})}^{2} - 12 (2^{x}) + 32 = 0 \Rightarrow (2^{x} - 4) (2^{x} - 8) = 0 \\ \Rightarrow 2^{x} = 2^{2}, 2^{3} \Rightarrow x = 2, 3 . \end{array}$
Clearly, $x \neq 2 .$ Therefore $x = 3$

If $\log_{5} 2, \log_{5} (2^{x} - 5)$ and $\log_{5} (2^{x} - 7 / 2)$ are in A.P., then x is equal to

Solution:

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