A) \[{{\log }_{e}}a\]
B) \[{{\log }_{e}}b\]
C) \[a\]
D) \[{{e}^{a}}\]
Correct Answer: C
Solution :
Given \[b={{\log }_{e}}(1+a)\Rightarrow 1+a={{e}^{b}}\] \[\Rightarrow \,\,\,1+a=1+\frac{b}{1!}+\frac{{{b}^{2}}}{2!}+....\]Þ \[a=b+\frac{{{b}^{2}}}{2\,!}+.....\]You need to login to perform this action.
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