How do negative powers and fractional powers make sense? [duplicate]

When $x \in \mathbb{N}$

You were probably taught that “exponentiation is repeated multiplication”:

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

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} & = 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] & = 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] & = 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] & = \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] & = \cos\theta + i \sin\theta \end{align}

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

$$e^{x+iy} = e^x \cdot e^{iy} = e^x (\cos{y} + i \sin{y})$$

This is easier to answer if you use $8$ instead of $2$. What does multiplying by $8^{1/3}$ mean? It means you multiply by $8$ one-third of one time, and that means you do something that, if done three times, amounts to multiplication by $8$. If you multiply by $2$ three times, you've multiplied by $8$. Therefore multiplying by $8$ one-third of one time is multiplying by $2$.

With $2^x$ instead of $8^x$, the idea is the same but the numbers are messy.

This leaves the question: What is $8^x$ if $x$ is not a rational number like $1/3$? The function $x\mapsto 8^x$ is monotone: as $x$ gets bigger, so does $8^x$. That means $8^x$ is bigger than $8^r$ when $r$ is any rational number less than $x$, and $8^x$ is less than $8^r$ when $r$ is any rational number bigger than $x$. That's enough to narrow down $8^x$ to just one number.

$f(x) = 2^x$ is the unique increasing real-valued function that satisfies $f(1)=2$ and $f(x+y) = f(x)f(y)$ for all $x,y \in \mathbb{R}$.

You can first define $2^r$ for rational numbers $r$: if $r = p/q$ where $p$ and $q$ are integers and $q > 0$, $2^r = (2^p)^{1/q}$ is the $q$'th root of $2^p$. It turns out that with this definition, $2^r$ is an increasing, continuous function of $r$. You can then define $2^x$ for any real number $x$ as the limit of $2^{r_n}$ for a sequence of rational numbers $r_n$ with limit $x$.