The expression 1+x log _e a+frac{x^2}{2 !}(lo | vedclass

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(B) We know that the Taylor series expansion for the exponential function is $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots$ Let $u = x \log

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$a^x$

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(B) We know that the Taylor series expansion for the exponential function is $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots$ Let $u = x \log _e a$. Then the given series becomes $1 + (x \log _e a) + \frac{(x \log _e a)^2}{2!} + \frac{(x \log _e a)^3}{3!} + \ldots$ This is equivalent to $e^{x \log _e a}$. Using the property of logarithms $n \log _e m = \log _e m^n$,we have $x \log _e a = \log _e a^x$. Therefore,$e^{\log _e a^x} = a^x$.

The expression $1+x \log _e a+\frac{x^2}{2 !}(\log _e a)^2+\frac{x^3}{3 !}(\log _e a)^3+\ldots$ for $a>0, x \in R$ is equal to:

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