A) \[x=0\]
B) \[x=1\]
C) \[x={{\log }_{2}}5\]
D) \[x={{\log }_{5}}2\]
Correct Answer: C
Solution :
[c] Let \[{{\log }_{10}}2,\,\,lo{{g}_{10}}\left( {{2}^{x}}-1 \right)\] and \[{{\log }_{10}}\left( {{2}^{x}}+3 \right)\]are in A.P \[\therefore \,\,\,\,2{{\log }_{10}}\left( {{2}^{x}}-1 \right)\] \[={{\log }_{10}}^{2}+{{\log }_{10}}\left( {{2}^{x}}+3 \right)\] \[\Rightarrow \,\,\,\,{{\log }_{10}}{{\left( {{2}^{x}}-1 \right)}^{2}}={{\log }_{10}}2\left( {{2}^{x}}+3 \right)\] \[\Rightarrow \,\,\,{{2}^{2x}}+1-{{2}^{x+1}}={{2.2}^{x}}+6\] \[\Rightarrow \,\,\,{{a}^{2}}+1-2a=2a+6\] where \[a={{2}^{x}}\]. \[\Rightarrow \,\,\,{{a}^{2}}-4a-5=0\] \[\Rightarrow \,\,\,a=5\] or \[a=-1\] \[{{2}^{x}}=5\Rightarrow \log 2=\log 5\] \[\Rightarrow \,\,\,\,x=\frac{\log 5}{\log 2}\Rightarrow x={{\log }_{2}}5\]
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