Finding the value of $a^x$ We have the series expansion $e^x = 1 + \frac{x}{1!}+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \frac{x^4}{4!}...\infty$. Is it possible to write $a^x$ in the similar form, where a is an integer and x could be any real value?

Yes, the standard definition is $a^x=e^{x\ln a}$, and you have already defined $e^y$ for any $y$.
The definition of $\ln a$ is $\ln a = \int_1^a \frac{1}{x} dx$.
I know it looks weird, but it turns out that this definition agrees with $a^n=a*a*...*a$ for any integer $n$, and also $a^{x+y}=a^xa^y$.
As a sanity check, look at $2^{0.001}$ on your calculator, and compare it with $\ln 2$.
