Expand a^x (a>0) in a power series in x, using known

Expand $a^x$ ($a>0$) in a power series in $x$, using known expansions, and indicate when it is valid.

Step 1 · Rewrite using base $e$

Since $a=e^{\ln a}$: $a^x=(e^{\ln a})^x=e^{x\ln a}$.

Step 2 · Substitute into the series for $e^t$

With $t=x\ln a$: $$ a^x = \sum_{n=0}^{\infty}\frac{(x\ln a)^n}{n!} = \sum_{n=0}^{\infty}\frac{(\ln a)^n}{n!}x^n = 1+(\ln a)x+\frac{(\ln a)^2}{2!}x^2+\cdots $$

Step 3 · Region of convergence

Since $e^t$ is entire and $t=x\ln a$ is linear in $x$, the series converges for all $x\in\mathbb R$.

Summary Table

Additional Notes

• When $a=e$, $\ln a=1$ and this reduces exactly to the ordinary exponential series; when $0
• Differentiating term by term recovers the expected identity $\frac{d}{dx}a^x=(\ln a)\cdot a^x$, confirming the series is self-consistent — the same style of derivative cross-check used throughout Problems 249–253.
• This is the general template underlying every exponential-type expansion in this batch: $e^x$ (Problem 249), $\sinh x$, $\cosh x$ (Problems 252–253), and $e^{-x^3}$ (Problem 259) are all special cases or close relatives of this same substitution-into-$e^t$ technique.