A) \[{{\log }_{2}}\left( \frac{1}{2} \right)\]
B) \[{{\log }_{2}}\left( \frac{5}{2} \right)\]
C) \[{{\log }_{2}}\left( \frac{1}{5} \right)\]
D) \[{{\log }_{2}}\left( \frac{2}{5} \right)\]
E) \[{{\log }_{2}}(5)\]
Correct Answer: D
Solution :
\[\because \] \[1,{{\log }_{4}}({{2}^{1-x}}+1),{{\log }_{2}}({{5.2}^{x}}+1)\]are in AP \[\therefore \] \[2{{\log }_{4}}({{2}^{1-x}}+1)={{\log }_{2}}({{5.2}^{x}}+1)+1\] \[\Rightarrow \] \[{{\log }_{2}}({{2}^{1-x}}+1)={{\log }_{2}}({{5.2}^{1+x}}+2)\] \[\Rightarrow \] \[{{2}^{1-x}}+1={{5.2}^{1+x}}+2\] \[\Rightarrow \] \[\frac{2}{t}=5.2t+1\] \[(Let\,t={{2}^{x}})\] \[\Rightarrow \] \[10{{t}^{2}}+t-2=0\] \[\Rightarrow \] \[10{{t}^{2}}+5t-4t-2=0\] \[\Rightarrow \] \[5t(2t+1)-2(2t+1)=0\] \[\Rightarrow \] \[t=\frac{2}{5},-\frac{1}{2}\] \[\left( \because t\ne -\frac{1}{2} \right)\] \[\therefore \] \[t=\frac{2}{5}\] \[\Rightarrow \] \[{{2}^{x}}=\frac{2}{5}\] \[\Rightarrow \] \[{{\log }_{2}}\left( \frac{2}{5} \right)=x\]
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