A) \[lo{{g}_{2}}5\]
B) \[1-lo{{g}_{2}}5\]
C) \[lo{{g}_{5}}2\]
D) None of these
Correct Answer: B
Solution :
The given numbers are in AP. \[\therefore \] \[2{{\log }_{4}}({{2}^{1-x}}+1)={{\log }_{2}}({{5.2}^{x}}+1)+1\] \[\Rightarrow \] \[2{{\log }_{2}}\left( \frac{2}{{{2}^{x}}}+1 \right)={{\log }_{2}}({{5.2}^{x}}+1)+{{\log }_{2}}2\] \[\Rightarrow \] \[\frac{2}{2}{{\log }_{2}}\left( \frac{2}{{{2}^{x}}}+1 \right)={{\log }_{2}}({{5.2}^{x}}+1)2\] \[\Rightarrow \] \[{{\log }_{2}}\left( \frac{2}{{{2}^{x}}}+1 \right)={{\log }_{2}}({{10.2}^{x}}+2)\] \[\Rightarrow \] \[\frac{2}{{{2}^{x}}}+1={{10.2}^{x}}+2\] \[\Rightarrow \] \[\frac{2}{y}+1=10y+2,\]where \[{{2}^{x}}=y\] \[\Rightarrow \] \[10{{y}^{2}}+y-2=0\Rightarrow (5y-2)(2y+1)=0\] \[\Rightarrow \] \[y=\frac{2}{5}\]or \[y=\frac{-1}{2}\Rightarrow {{2}^{x}}=\frac{2}{5}\]or\[{{2}^{x}}=\frac{-1}{2}\] \[\Rightarrow \] \[x={{\log }_{2}}\left( \frac{2}{5} \right)\] \[\Rightarrow \] \[x={{\log }_{2}}2-{{\log }_{2}}5\] \[\Rightarrow \] \[x=1-{{\log }_{2}}5\]
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