What does $2^x$ really mean when $x$ is not an integer? We all know that $2^5$ means $2\times 2\times 2\times 2\times 2 = 32$, but what does $2^\pi$ mean? How is it possible to calculate that without using a calculator? I am really curious about this, so please let me know what you think.
When $x \in \mathbb{N}$

You were probably taught that &ldquo;exponentiation is repeated multiplication&rdquo;:

$$b^x = \underbrace{b\times b\times b\times\cdots\times b}_{x\text{ times}}$$

From this simple definition, you can observe two properties:


$b^{x+y} = b^x \cdot b^y$
$b^{xy} = \left(b^x\right)^y $


For example:


$2^{3+4} = 2^7 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = (2 \cdot 2 \cdot 2) \cdot (2 \cdot 2 \cdot 2 \cdot 2) = 2^3 \cdot 2^4$
$2^{3 \cdot 4} = 2^{12} = 2^3 \cdot 2^3 \cdot 2^3 \cdot 2^3 = \left(2^3\right)^4$


We can then definite exponentation over more general sets of numbers in a way that these two properties continue to hold.

When $x \in \mathbb{Z}$

From the above rule for addition of exponents, we obtain a rule for subtraction of exponents:  $b^{x - y} = {b^x \over b^y}$, because then $b^{(x - y) + y} = b^{x-y} \cdot b^y = {b^x \over b^y} \cdot b^y = b^x$ as expected.  This lets us expand the domain of exponents to include zero and negative integers:

$$b^0 = b^{y-y} = {b^y \over b^y} = 1,\; b \ne 0$$
$$b^{-y} = b^{0-y} = {b^0 \over b^y} = {1 \over b^y},\; b \ne 0$$

When $x \in \mathbb{Q}$

If you assume that the multiplicate property of exponents holds for rationals, then $\left(b^{1 \over n}\right)^n = b^{{1 \over n} \cdot n} = b^1 = b$.  So $b^{1 \over n}$ is a number whose $n$th power is $b$.  In other words,

$$b^{1 \over n} = \sqrt[n]{b},\; b \ge 0$$

And $b^{m \over n} = \left(b^{1 \over n}\right)^m = (\sqrt[n]{b})^m$.

For example, $4096^{5/12} = \left(\sqrt[12]{4096}\right)^5 = 2^5 = 32$.

When $x \in \mathbb{R}$

I still haven't answered your question of what $2^\pi$ means.  But at this point, we can calculate $2^x$ for $x$ aribitrarily close to $\pi$.


$2^3$ = 8
$2^{3.1} = 2^{31/10} = \sqrt[10]{2^{31}} \approx 8.574187700290345$
$2^{3.14} = 2^{314/100} = \sqrt[100]{2^{314}} \approx 8.815240927012887$
$2^{3.141} = 2^{3141/1000} = \sqrt[1000]{2^{3141}} \approx 8.821353304551304$
$2^{3.1415} = 2^{31415/1000} = \sqrt[10000]{2^{31415}} \approx 8.824411082479122$
$2^{3.14159} = 2^{314159/10000} = \sqrt[100000]{2^{314159}} \approx 8.824961595059897$


As $x$ approaches $\pi$, $2^x$ approaches a limit, which is approximately $8.824977827076287$.  For the sake of making $2^x$ continuous, we define $2^{\pi}$ to be equal to this limit.

(Note that there's nothing special about decimal fractions.  I could have used the sequence $[3, {22 \over 7}, {333 \over 106}, {355 \over 113}, \ldots ]$ of best rational approximations, but that would have been less obvious.)

However, taking the trillionth root of huge powers of a number isn't very practical for calculation.  A more useful method is to use logarithms.

$\log_c y$ is defined as the number $x$ such that $c^x = y$.  From the two basic properties of exponentation, you can obtain the identities:


$\log_c (ab) = \log_c a + \log_c b$
$\log_c (b^x) = x \cdot \log_c b$


And from the latter, you get $$b^x = c^{x \cdot \log_c b}.$$ This means that if you have an exponential and logarithm function for one value of $c$, you can calculate them for any value for $b$.

Typical choice of $c$ are:


2, for convenience in working with computers
10, the base of our number system, giving "common logarithms"
$e \approx 2.718281828459045$, the base of the "natural logarithm" ($\ln$), for its convenient properties in calculus.


So, if you wanted to calculate $2^{\pi}$, you'd actually calculate $10^{\pi \cdot \log_{10} 2}$ or $e^{\pi \cdot \ln 2}$.  And that would typically be done with the assistance of a logarithm table or a slide rule.

When $x \in \mathbb{C}$

In Calculus, you'll learn about Taylor series, and the well-known ones for $e^x$, sine and cosine:


$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \dots = \sum_{k=0}^{\infty} \frac{x^k}{k!}$
$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}$
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \frac{x^8}{40320} - \cdots = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k+1)!}$


What happens when you plug $x = i \theta$ into the Taylor series for $e^x$?

\begin{align}
e^{i \theta} &amp; = 1 + i \theta + \frac{(i \theta)^2}{2} + \frac{(i \theta)^3}{6} + \frac{(i \theta)^4}{24} + \frac{(i \theta)^5}{120} + \frac{(i \theta)^6}{720} + \frac{(i \theta)^7}{5040} + \frac{(i \theta)^8}{40320} + \cdots \\[10pt]
&amp; = 1 + i \theta + i^2 \frac{\theta^2}{2} + i^3 \frac{\theta^3}{6} + i^4 \frac{\theta^4}{24} + i^5 \frac{\theta^5}{120} + i^6 \frac{\theta^6}{720} + i^7 \frac{\theta^7}{5040} + i^8 \frac{\theta^8}{40320} + \cdots \\[10pt]
&amp; = 1 + i \theta - \frac{\theta^2}{2} - i \frac{\theta^3}{6} + \frac{\theta^4}{24} + i \frac{\theta^5}{120} - \frac{\theta^6}{720} - i \frac{\theta^7}{5040} + \frac{\theta^8}{40320} + \cdots \\[10pt]
&amp; = \left( 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \frac{\theta^6}{720} + \frac{\theta^8}{40320} - \dots\right) + i \left(\theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \frac{\theta^7}{5040} + \cdots \right) \\[10pt]
&amp; = \cos\theta + i \sin\theta
\end{align}

This is called Euler's formula, and it lets us extend exponentiation to the complex numbers:

$$e^{x+iy} = e^x \cdot e^{iy} = e^x (\cos{y} + i \sin{y})$$
Euler's identity would be another use of exponents outside of integers as there are some complex numbers used in the identity.  Euler's formula explains how to evaluate such values which does help in some cases to evaluate the function.