# 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.

Thanks!

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Good question! Here's one intuitive answer: if you know that a population of bacteria doubles in size every hour, then under some reasonable assumptions $2^{\pi}$ is approximately how much larger the population is after $\pi$ hours. – Qiaochu Yuan Apr 16 '12 at 21:54
Much like multiplication is not repeated addition (continued here), exponentiation is not repeated multiplication. – Asaf Karagila Apr 16 '12 at 21:58
The links work for me. @AsafKaragila: I have to admit I was unable to get Devlins point in the first column. Apparently his point is that the real numbers are the natural all purpose number system and everything should be done starting from the reals. The bottom down approach. Of course the bottom up approach as, say, developed in Landau's little book does work. And I don't see how children will ever learn about addition without telling them it is...repeated counting. The natural numbers are special! – Michael Greinecker Apr 16 '12 at 23:01
@AsafKaragila pedagogically Delvin's advice for teachers is very bad. Yes, multiplication is repeated addition (that is why it was first used, and that is how primary students ought to understand it): it's just that mathematicians have extended this idea to apply also to non-integer numbers. Compare: factorial function/gamma function. – Ronald Apr 16 '12 at 23:15
@AsafKaragila: Would it be better to say that multiplication is generalized repeated addition? Or am I missing the point here? (Interestingly, if you look at it this way it seems a large part of modern mathematics is just generalized addition: measure theory, algebra, analysis, cardinal/ordinal arithmetic ... Maybe we should say that mathematics is generalized addition =)) – Dejan Govc Apr 17 '12 at 10:11

This is easier to answer if you use $8$ instead of $2$. What does $8^{1/3}$ 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.

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Thanks for the great answer. This really simplified it for me! – David Apr 16 '12 at 22:10
This has got to be the best explanation for a fractional power I have ever seen. – Apoorv Khurasia Apr 17 '12 at 1:18

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.

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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 $x$.

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Additionally, $2^{-p}$ can be interpreted as $(2^{-1})^p$, which leaves the problem of interpreting $2^{-1}$. If we want the exponent laws to hold, we should have $(2^1)(2^{-1})=2^0=1$, which rearranges to $2^{-1}=\frac1{2^1}$... – J. M. Apr 17 '12 at 7:32

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

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+1, that's a good argument, but I think perhaps you should develop the idea a little more. – JMCF125 Feb 3 '14 at 20:02
Agree with @JMCF125 , you should expand on why this happens, why it is unique. I like your idea. – Sawarnik Feb 26 '14 at 16:24

Think of it this way:

$2^5 = 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$

because of the property that $a^b \cdot a^c = a^{b+c}$

Now examine $2^{\pi}$

$2^{\pi} = 2^{3.1415926535...} = 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^{.1415926535...} \approx 8 \cdot \sqrt[7]2 \approx 8.8249778...$

For any power (aside from zero), if it is negative flip the term over (reciprocal), take the whole terms out as powers, and think of any remaining decimals as 'roots.'

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# When $x \in \mathbb{N}$

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

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

From this simple definition, you can observe two properties:

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

For example:

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

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

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

# 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 $e^x$?

$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= 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= 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= (1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \frac{\theta^6}{720} + \frac{\theta^8}{40320} - \dots) + i (\theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \frac{\theta^7}{5040} + \cdots)= \cos\theta + i \sin\theta $$This is called Euler's formula, and it lets us extend exponentiation to the complex numbers:$$e^{x+iy} = e^x \cdot e^{iy} = e^x (\cos{y} + i \sin{y})$$- Could you please expand this to when x \in \mathbb{C}? – Anton Golov Apr 19 '12 at 21:32 @Dan, great compendium! – Pedro Tamaroff Apr 20 '12 at 2:32 @AntonGolov: OK, but I'm not doing quaternions. – Dan Apr 20 '12 at 2:33 @Dan, but what about other number systems (other than quaterions) eg field extensions, ideal domains etc? – alancalvitti Jan 8 '13 at 23:57 Excellent answer. – Jp McCarthy Oct 17 '13 at 19:31 From the "formal mathematics" point of view, one could argue that you are asking the wrong question. The right question to ask may be "What is \log 2?", which in turn leads to the question "What is \log x?" Let me explain what I mean, and why these questions are related to your question. Formally, we define \log x as an integral:$$\log x := \int_1^x \frac{1}{t} dt$$so that \log 2 is just the area under f(t)=\frac{1}{t} between t=1 and t=2. From this definition, one can prove all the properties that one expects from \log x. For instance: • The domain of \log x is (0,\infty), • The function \log x is continuous in its domain, • The function \log x is differentiable in its domain, • The function \log x is strictly increasing in its domain, • \log(xy)=\log x + \log y, for any x,y>0, • \log(x^y)=y\log x, for any x>0 and any y\in\mathbb{R}. Since \log x is continuous and strictly increasing in its domain, the function \log x is invertible, and we define e^x as the inverse function of \log x, so that e^{\log x} = x, for any x>0, and \log e^x = x, for any real number x. Now that we have defined \log x (in terms of areas under 1/t) and we have defined e^x (as the inverse function to \log x), we can talk about 2^5:$$2^5 = e^{\log 2^5} = (e^{5\log 2})$$and the same works for any real number \alpha:$$2^\alpha = e^{\log 2^\alpha} = e^{\alpha\log 2}.$$If we use the fact that e^x is the inverse function of \log x, and the definition of \log in terms of areas, we reach the conclusion that: • " 2^\alpha " is the number x such that the area between t=1 and t=x under \frac{1}{t} is precisely$$\alpha \log 2 = \alpha\cdot (\text{the area under } 1/t \text{ between } t=1 \text{ and } t=2).$$- with that way you can even easily extend the definition to complex numbers (the branch cut becomes irrelevant by the exponentiation adding a power of one) – Tobias Kienzler Apr 18 '12 at 6:37 If this helps: For all n:$$2^n=e^{n\log 2}$$This is a smooth function that is defined everywhere. Another way to think about this (in a more straightforward manner then others described): We know$$a^{b+c}=a^ba^c$$Then say, for example, b=c=1/2. Then we have:$$a^{1}=a=a^{1/2}a^{1/2}$$Thus$a^{1/2}=\sqrt{a}$is a number that equals$a$when multiplied by itself. Now we can find the value of (for some p and q),$a^{p/q}$. We know:$(a^x)^y=a^{xy}$thus$(a^{p/q})^{q/p}=a^1=a$Other exponents may be derived similarly. - We assume that$\mathbb{R}$is constructed using equivalence classes of rational Cauchy-sequences and$\alpha^\frac{1}{n}$is defined as the unique non-negative solution of$x^n = \alpha$, where$\alpha$is non-negative. For rational$a=\frac{p}{q}$,$p,q \in \mathbb{Z}$,$2^a$is defined by$(2^\frac{1}{q})^p$. We first show that$2^a$from$\mathbb{Q}$to$\mathbb{R}$is a strictly increasing function. Take$a,b\in\mathbb{Q}$,$a<b$. Then$a = \frac{p_1}{q_1}$,$q_1 > 0$and$b = \frac{p_2}{q_2}$,$q_2 > 0$, where$p_i,q_i\in \mathbb{Z}$. The following lines are equivalent \begin{eqnarray} 2^a & < & 2^b \\ 2^\frac{p_1}{q_1} & < & 2^\frac{p_2}{q_2} \\ (2^\frac{1}{q_1 q_2})^{p_1 q_2} & < & (2^\frac{1}{q_1 q_2})^{p_2 q_1} \\ p_1 q_2 & < & p_2 q_1 \\ \frac{p_1}{q_1} & < & \frac{p_2}{q_2} \\ a & < & b \ . \end{eqnarray} The fourth line is the definition of the ordering relation$<$in$\mathbb{Q}$. Hence$2^a$is strictly increasing. Take now any Cauchy-sequence$\{a_n\}_{n = 0}^\infty \subset \mathbb{Q}$. Because$a_n$is a Cauchy-sequence, it is bounded and there is an integer$m$s.t.$a_{n_k} < m$. Assume that$2^\frac{1}{j}$does not converge to$1$as$j \rightarrow \infty$. The sequence is decreasing and bounded below. It has a limit. Note that the limit cannot be reached because the sequence is strictly decreasing. Let the limit be$\beta > 1$. Then we have$2^\frac{1}{j} > \beta$for every$j \in \mathbb{Z}^+$. We can now estimate \begin{eqnarray} 2 = x_j^j > \beta^j \rightarrow \infty \ , \end{eqnarray} that is a contradiction. Hence $$\lim_{j \rightarrow \infty} 2^\frac{1}{j} = 1 \ .$$ We next estimate the sequence$a_n$. Choose$\epsilon > 0$. Choose$j \in \mathbb{Z}^+$s.t.$ 2^m (2^\frac{1}{j} - 1) < \epsilon$. Choose$N \in \mathbb{N}$s.t.$k,l>N$implies$|a_k-a_l| < \frac{1}{j}$. We estimate \begin{eqnarray} |2^{a_k}-2^{a_l}| & = & 0 < \epsilon \ \ , \ a_k = a_l \ , \\ |2^{a_k}-2^{a_l}| & = & 2^{a_l} (2^{a_k-a_l} - 1) < 2^m (2^\frac{1}{j}-1) < \epsilon \ \ , \ a_k > a_l \ , \\ |2^{a_k}-2^{a_l}| & = & 2^{a_l}-2^{a_k} = 2^{a_k} (2^{a_l-a_k} - 1) < 2^m (2^\frac{1}{j}-1) < \epsilon \ \ , \ a_k < a_l \ . \end{eqnarray} Hence$2^{a_k}$is a Cauchy-sequence and it converges to a limit. Continuity of$2^a$can be shown as follows. Assume that$a_n$converges to a limit$a\in\mathbb{Q}$. Then replace$a_l$by$a\$ in estimates and obtain the definition of convergence.

We define $$2^x = \lim_{n \rightarrow \infty} 2^{a_n} \ .$$

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