A) \[5\]
B) \[3\]
C) \[{{\log }_{3}}2\]
D) \[{{\log }_{2}}5\]
Correct Answer: D
Solution :
If \[\log 2,\,\log \,({{2}^{x}}-1),\] \[\log \,({{2}^{x}}+3)\] are in A.P, then \[2\log ({{2}^{x}}-1)=log2+log({{2}^{x}}+3)\] \[\Rightarrow \,\,{{({{2}^{x}}-1)}^{2}}=2({{2}^{x}}+3)\] \[\Rightarrow \,\,{{y}^{2}}-2y+1=2y+6\] (Put \[{{2}^{x}}=y\]) \[\Rightarrow \,\,{{y}^{2}}-4y-5=0\Rightarrow (y+1)\,(y-5)=0\] \[\Rightarrow \,\,{{2}^{x}}=-1\] or \[{{2}^{x}}=5\] But \[{{2}^{x}}\ne -1\forall x\] \[\therefore \,\,\,{{2}^{x}}=5\Rightarrow \log \,{{2}^{x}}=\log 5\] \[\Rightarrow \,\,x=\frac{\log 5}{\log 2}\Rightarrow x={{\log }_{2}}5\]
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