A) \[1,\,\,\frac{1}{2}\]
B) \[1,\,\,\frac{1}{3}\]
C) \[1,\,\,\frac{3}{2}\]
D) None of these
Correct Answer: D
Solution :
Since, it is an\[AP\]. \[\therefore \] \[{{\log }_{3}}({{2}^{x}}-5)=\frac{{{\log }_{3}}2+{{\log }_{3}}\left( {{2}^{x}}-\frac{7}{2} \right)}{2}\] \[\Rightarrow \] \[2{{\log }_{3}}({{2}^{x}}-5)={{\log }_{3}}\left( 2\cdot \left( {{2}^{x}}-\frac{7}{2} \right) \right)\] \[\Rightarrow \] \[2{{\log }_{3}}{{({{2}^{x}}-5)}^{2}}={{2}^{x+1}}-7\] \[\Rightarrow \] \[{{2}^{2x}}-12\cdot {{2}^{x}}-32=0\] \[\Rightarrow \] \[x=2,\,\,3\] But\[x=2,\,\,{{\log }_{3}}({{2}^{x}}-5)\]is not defined.
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