If log52,log5(2x−5) and log5(2x−72) are in A.P. Then x=

log52,log5(2x−5) and log5(2x−72) are in AP ⇒2log5(2x−5)=log52+log2(2x−72) ⇒(2x−5)2=2(2x−72)⇒(2x)2−12(2x)+32=0

If log52,log5(2x−5) and log5(2x−72) are in A.P. Then x=

log52,log5(2x−5) and log5(2x−72) are in AP ⇒2log5(2x−5)=log52+log2(2x−72) ⇒(2x−5)2=2(2x−72)⇒(2x)2−12(2x)+32=0

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

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

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

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