A) \[x=a\]
B) \[x={{a}^{a}}\]
C) \[x={{a}^{-1/a}}\]
D) \[x={{a}^{1/a}}\]
Correct Answer: D
Solution :
\[{{\log }_{a}}x+2{{\log }_{a}}x+.......+a{{\log }_{a}}x=\frac{a+1}{2}\] \[\Rightarrow \] \[{{\log }_{a}}x(1+2+........+a)=\frac{a+1}{2}\] \[\Rightarrow \] \[{{\log }_{a}}x.\frac{a(a+1)}{2}=\frac{a+1}{2}\]\[\Rightarrow \]\[({{10}^{12}}+{{10}^{11}}+......+1)\].
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