A) \[{{a}^{x}}\]
B) x
C) \[{{a}^{{{\log }_{a}}x}}\]
D) a
Correct Answer: A
Solution :
Series = \[1+x{{\log }_{e}}a+\frac{{{x}^{2}}}{2!}{{\left[ {{\log }_{e}}a \right]}^{2}}+\frac{{{x}^{3}}}{3!}{{[{{\log }_{e}}a]}^{3}}+...\] \[={{e}^{x{{\log }_{e}}a}}={{e}^{{{\log }_{e}}{{a}^{x}}}}={{a}^{x}}\].
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