If log3 2 log3 (2^x – 5) log3 (2^x – 7/2) are in A.P then x is 

Correct option is (A) an odd positive integer

Given that \(log_3\,2,\,log_3\,(2^x-5)\;and\;log_3\,(2^x-\frac{7}{2})\) are in A.P.

\(\therefore\) \(log_3\,(2^x-5)=\frac{log_3\,2+log_3\,(2^x-\frac{7}{2})}2\)

\(\Rightarrow\) \(2\,log_3\,(2^x-5)=log_3\,2+log_3\,(2^x-\frac{7}{2})\)

\(\Rightarrow\) \(log_3\,(2^x-5)^2=log_3\,\left(2(2^x-\frac{7}{2})\right)\)

\(\therefore\) \((2^x-5)^2=2\times2^x-7\)      (By comparing)

\(\Rightarrow\) \(2^{2x}-10.2^x+25=2.2^x-7\)

\(\Rightarrow\) \(2^{2x}-12.2^x+25+7=0\)

\(\Rightarrow\) \(2^{2x}-12.2^x+32=0\)

\(\Rightarrow\) \(2^{2x}-8.2^x-4.2^x+32=0\)

\(\Rightarrow2^x(2^x-8)-4(2^x-8)=0\)

\(\Rightarrow(2^x-8)(2^x-4)=0\)

\(\Rightarrow\) \(2^x\) - 8 = 0 or \(2^x\) - 4 = 0

\(\Rightarrow2^x=8=2^3\) or \(2^x=4=2^2\)

\(\Rightarrow\) x = 3 or x = 2

\(\therefore x\neq2\) because if x = 2 then \(2^x-5=2^2-5\) = 4 - 5 = -1 which can not be a domain for log function.

\(\therefore\) x = 3 which is an odd positive integer.

Correct option is (A) an odd positive integer