# What does 2×2^x really mean when xx 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 “exponentiation is repeated multiplication”:

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:

# 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,

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

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

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

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 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 $e^x$?

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