A) \[1,\ \frac{1}{2}\]
B) \[1,\ \frac{1}{3}\]
C) \[1,\ \frac{3}{2}\]
D) None of these
Correct Answer: D
Solution :
\[{{\log }_{3}}2,\ {{\log }_{3}}({{2}^{x}}-5)\] and \[{{\log }_{3}}\left( {{2}^{x}}-\frac{7}{2} \right)\] are in A.P. \[\Rightarrow \]\[2{{\log }_{3}}({{2}^{x}}-5)={{\log }_{3}}\left[ (2)\,\left( {{2}^{x}}-\frac{7}{2} \right) \right]\] \[\Rightarrow \] \[{{({{2}^{x}}-5)}^{2}}={{2}^{x+1}}-7\]\[\Rightarrow \]\[{{2}^{2x}}-12\ .\ {{2}^{x}}-32=0\] \[\Rightarrow \] \[x=2,\ 3\] But \[x=2\] does not hold, hence\[x=3\].
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