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Prove that ax = ∑ ((log a)n / n!) xn

Assuming that a^x has a power-series representation in terms of powers of x, verify that it has the form

    \[ a^x = \sum_{n=0}^{\infty} \frac{(\log a)^n}{n!} x^n, \qquad a > 0 \]

valid for all real x.


From the definition of the exponential we have

    \[ a^x = e^{x \log a}. \]

We also know the series expansion for e^x,

    \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \]

Therefore,

    \begin{align*}  a^x &= e^{x \log a} \\  &= \sum_{n=0}^{\infty} \frac{(x \log a)^n}{n!} \\[9pt]  &= \sum_{n=0}^{\infty} \frac{(\log a)^n}{n!} x^n. \end{align*}

This converges everywhere x \log a is defined (since the series expansion of e converges for all real x). Hence, this is valid for all x and all a > 0 (since that is where \log a is defined).

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