A) \[{{\log }_{b}}a\]
B) \[{{\log }_{a}}b\]
C) \[{{\log }_{e}}a-{{\log }_{e}}b\]
D) \[{{\log }_{e}}a+{{\log }_{e}}b\]
Correct Answer: A
Solution :
\[\frac{(a-1)-\frac{{{(a-1)}^{2}}}{2}+\frac{{{(a-1)}^{3}}}{3}-.......\infty }{(b-1)-\frac{{{(b-1)}^{2}}}{2}+\frac{{{(b-1)}^{3}}}{3}-......\infty }\] \[=\frac{{{\log }_{e}}(1+\overline{a-1})}{{{\log }_{e}}(1+\overline{b-1})}=\frac{{{\log }_{e}}a}{{{\log }_{e}}b}={{\log }_{b}}a\].
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