Is there an intuitive definition of exponentiation?

In elementary school, we learned that

$$ a^b = a \cdot a \cdot a \cdot a \cdots (b\ \textrm{ times}) $$ where $b$ is an integer.

Then later on this was expanded to include rational exponents, so that

$$ a^{\frac{b}{c}} = \sqrt[c]{a^b} $$

From there we could evaluate decimal exponents like $4^{3.24}$ by first converting to a fraction.

However, even after learning Euler's Identity, I feel as though there is no discussion on what exponentiation really means. The definitions I found are either overly simplistic or unhelpfully complex. Once we stray from the land of rational powers into real powers in general, is there an intuitive definition or explanation of exponentiation?

I am thinking along the lines of, for example, $2^\pi$ or $3^{\sqrt2}$ (or any other irrational power, really). What does this mean? Or, is there no real-world relationship?

To draw a parallel to multiplication:
If we consider the expression $e\cdot \sqrt5$, I could tell you that this represents the area of a rectangle with side lengths $e$ cm and $\sqrt5$ cm. Or maybe $e \cdot \pi$ is the cost of $\pi$ kg of material that costs $e$ dollars per kg. Of course these quantities would not be exact, but the underlying intuition does not break down. The idea of repeated addition still holds, just that fractional parts of terms, rather than the entire number, are being added.

So does such an intuition for exponentiation exist? Or is this one of the many things we must accept with proof but not understanding?

This question stems from trying to understand complex exponents including Euler's identity and $2^i$, but I realized that we must first understand reals before moving on the complex numbers.

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    $\begingroup$ Yes, that I know. But I am also asking about irrational algebraic numbers such as roots, and also what exponents with transcendental/roots mean. $\endgroup$
    – baum
    Commented Mar 7, 2014 at 0:27
  • 11
    $\begingroup$ My answer is quite long, so I'll summarize here: yes, there is a very intuitive definition of real exponentiation. For natural, integer, and rational powers it's exactly what you'd expect; for arbitrary real powers it is the unique continuous extension of the usual exponential to the real numbers. $\endgroup$ Commented Mar 7, 2014 at 1:02
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    $\begingroup$ Exponentiation is related to growth that is always proportional to how large the quantity has grown so far. You can have different starts and rates but you can imagine the growth as happening gradually and continuously, and imagining how that would interpolate is how you extend to reals. $\endgroup$
    – Patashu
    Commented Mar 7, 2014 at 2:33
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    $\begingroup$ My general philosophy is that you what something really means is nothing more than what you can do with it and how it can be used to solve problems and explain ideas. $\endgroup$
    – user14972
    Commented Mar 7, 2014 at 8:23
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    $\begingroup$ It all depends on what you mean by 'mean'. If it is simply a matter of "I don't get it" despite being able to do the manipulations, one could say that there is no intention of you to 'get it', it just a mindless preservation of the rules for basic exponentiation (which I think you do 'get') to allow some calculation. If 'mean' means "I don't see a quick generalization that encompasses integers and reals" then the best generalization is to complex numbers and scaling rotation. (hmm, that doesn't explain it for reals though...OK stick with the calculus/derivative equals itself.) $\endgroup$
    – Mitch
    Commented Mar 7, 2014 at 13:42

13 Answers 13


My chief understanding of the exponential and the logarithm come from Spivak's wonderful book Calculus. He devotes a chapter to the definitions of both.

Think of exponentiation as some abstract operation $f_a$ ($a$ is just some index, but you'll see why it's there) that takes a natural number $n$ and spits out a new number $f_a(n)$. You should think of $f_a(n) = a^n$.

To match our usual notion of exponentiation, we want it to satisfy a few rules, most importantly $f_a(n+m) = f_a(n)f_a(m)$. Like how $a^{n+m} = a^na^m$.

Now, we can extend this operation to the negative integers using this rule: take $f_a(-n)$ to be $1/f_a(n)$. then $f_a(0) = f_a(n-n) = f_a(n)f_a(-n) = 1$, like how $a^0=1$.

Then we can extend the operation to the rational numbers, by taking $f_a(n/m) = \sqrt[m]{f_a(n)}$. Like how $a^{n/m} = \sqrt[m]{a^n}$.

Now, from here we can look to extend $f_a$ to the real numbers. This takes more work than what's happened up to now. The idea is that we want $f_a$ to satisfy the basic property of exponentiation: $f_a(x+y)=f_a(x)f_a(y)$. This way we know it agrees with usual exponentiation for natural numbers, integers, and rational numbers. But there are a million ways to extend $f_a$ while preserving this property, so how do we choose?

Answer: Require $f_a$ to be continuous.

This way, we also have a way to evaluate $f_a(x)$ for any real number $x$: take a sequence of rational numbers $x_n$ converging to $x$, then $f_a(x)$ is $\lim_{n\to\infty} f_a(x_n)$. This seems like a pretty reasonable property to require!

Now, actually constructing a function that does this is hard. It turns out it's easier to define its inverse function, the logarithm $\log(z)$, which is the area under the curve $y=1/x$ from $1$ to $z$ for $0<z<\infty$. Once you've defined the logarithm, you can define its inverse $\exp(z) = e^z$. You can then prove that it has all the properties of the exponential that we wanted, namely continuity and $\exp(x+y)=\exp(x)\exp(y)$. From here you can change the base of the exponential: $a^x = (e^{\log a})^x = e^{x\log a}$.

To conclude: the real exponential function $\exp$ is defined (in fact uniquely) to be a continuous function $\mathbb{R}\to\mathbb{R}$ satisfying the identity $\exp(x+y)=\exp(x)\exp(y)$ for all real $x$ and $y$. One way to interpret it for real numbers is as a limit of exponentiating by rational approximations. Its inverse, the logarithm, can similarly be justified.

Finally, de Moivre's formula $e^{ix} = \cos(x)+i\sin(x)$ is what happens when you take the Taylor series expansion of $e^x$ and formally use it as its definition in the complex plane. This is more removed from intuition; it's really a bit of formal mathematical symbol-pushing.

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    $\begingroup$ This is a good definition of exponentiation, or even a good motivation for the definition, but I think it falls short of actually giving an intuition for what it really "means". $\endgroup$ Commented Mar 7, 2014 at 6:18
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    $\begingroup$ +1, but I'd say that first defining $\ln x=\int_{1}^x \frac1t\,dt$ is just one possibility. Depending on what you already know it may not be more difficult to directly define $\exp x=\sum_{k=0}^\infty\frac1{k!}x^k$. $\endgroup$
    – Carsten S
    Commented Mar 7, 2014 at 8:28
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    $\begingroup$ @ShreevatsaR For many mathematical objects, at some point you have to give up on exact analogies to familiar cases. Exponentiation by a real number is one of them - the strict analogy to "multiplying by itself $n$ times" or taking "$n$-th roots" quickly becomes a point of confusion. At this point I think strong motivation is the best substitute for exact analogy. And I still think this is a pretty intuitive way to think about the exponential: rational exponentiation is an intuitive process, and making it continuous is reasonably intuitive to a freshman calculus student (fill in the holes!). $\endgroup$ Commented Mar 7, 2014 at 10:41
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    $\begingroup$ For exp to be unique, we need some other additional criterion, as the homomorphy + continuity property is valid for every $f(x) = a^x$ function, not just for $a = e$. $\endgroup$ Commented Mar 8, 2014 at 9:45
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    $\begingroup$ @PaŭloEbermann Thanks for that comment, that's important. Here though, I'd like to think about extending the exponential for a particular base $a$, and in that case we have uniqueness of the extension. $\endgroup$ Commented Mar 13, 2014 at 5:04

$2^\pi$ or $3^\sqrt2$ (or any other irrational power, really). What does this mean?

$$a^\pi=a^{3.1415\ldots}=a^{3\ +\ 0.1\ +\ 0.04\ +\ 0.001\ +\ 0.0005\ +\ \cdots}=a^3\cdot a^{0.1}\cdot a^{0.04}\cdot a^{0.001}\cdot a^{0.0005}\cdots$$

$$a^\sqrt2=a^{1.4142\ldots}=a^{1\ +\ 0.4\ +\ 0.01\ +\ 0.004\ +\ 0.0002\ +\ \cdots}=a^1\cdot a^{0.4}\cdot a^{0.01}\cdot a^{0.004}\cdot a^{0.0002}\cdots$$

It is obvious that the general factor of this infinite product tends towards $a^0=1$. Convergence then follows from the fact that each single decimal digit is in between $0$ and $9$, meaning that $\displaystyle\prod_ka^{^{\tfrac{d_k}{10^k}}}$ is in between $\displaystyle\prod_ka^{^{\tfrac0{10^k}}}=1$, and $\displaystyle\prod_ka^{^{\tfrac9{10^k}}}=a^{10^{-n}}$, where $n$ is the number of digits of $[a]$.

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    $\begingroup$ isn't $a^{\pi} < max(a^3,a^4)$ enough to say it's convergent (since it's also increasing with your decomposition) $\endgroup$
    – caub
    Commented Mar 7, 2014 at 20:47
  • $\begingroup$ Since it's both bounded and monotonous, it converges, yes. $\endgroup$
    – Lucian
    Commented Mar 7, 2014 at 20:51
  • $\begingroup$ It might be good to note that you're essentially requiring continuity to extend the "intuitive" definition, just as in kigen's answer $\endgroup$
    – M Turgeon
    Commented Mar 9, 2014 at 19:04

One reason your intuition might break down here is that exponentiation - even integer exponentiation - is inherently dimensionless. If a square has area of $5\mathrm{cm}^2$ then you can talk about its sides having length $\sqrt{5\mathrm{cm}^2} = \sqrt{5}\mathrm{cm}$, but there are no units for $x$ that give $2^x$ a sensible set of units — try to figure out the units in the formula $2^{a+b} = 2^a2^b$. That said, this doesn't mean that there can't be understanding; there are many different ways of understanding something. But the physical understanding is going to be difficult.

OTOH, 'difficult' isn't necessarily 'impossible'; one reasonable approach to understanding the exponential is via the classic rate-change model. If I have $1$kg of paramecia and they're doubling their mass every minute, what will be the mass of the paramecia after $2.35$ minutes? Note that this gets around the dimension barrier I mentioned in the previous paragraph, because the quantity being exponentiated is dimensionless; it has dimensions of (time)/(time) (with the numerator being the elapsed time and the denominator being the doubling time).

With that said, though, this approach is going to be effectively impossible to extend to complex exponents (what would $5+2i$ time intervals be?), and my hearty recommendation would be to try and get away from the physical understanding in your head; don't map the exponential to something that you have 'real-world' experience with, but instead try to understand it on its own terms and understand its properties directly.


To draw a parallel to multiplication:
If we consider the expression $e\cdot \sqrt5$, I could tell you that this represents the area of a rectangle with side lengths $e$ cm and $\sqrt5$ cm. Or maybe $e \cdot \pi$ is the cost of $\pi$ kg of material that costs $e$ dollars per kg. Of course these quantities would not be exact, but the underlying intuition does not break down. The idea of repeated addition still holds, just that fractional parts of terms, rather than the entire number, are being added.

So does such an intuition for exponentiation exist?

Of course! For this we'll need an agar plate and some bacteria.

enter image description here

Illustrating exponents for whole numbers

You have a (very large) agar plate into which you add three bacteria. The bacteria reproduce and will double precisely every hour.

After $h$ hours you will have $3 \times 2^{h}$ bacteria.

I am thinking along the lines of, for example, $2\pi$ or $3^\sqrt{2}$ (or any other irrational power, really). What does this mean? Or, is there no real-world relationship?

There is indeed!

Illustrating exponents for real numbers

In fact, our last example was not very real-world. We assumed that the number of bacteria would instantaneously double precisely each time the big hand of the clock reaches 12. But bacteria are not so well-behaved.

Instead let's take a more reasonable assumption: that scientists have done experiments, monitoring the bacteria precisely when the big hand of the clock reaches twelve. Each time they observe precisely twice as many bacteria as the last experiment. But the scientists also know that the bacteria can increment continuously: any time in between.

Now let's say we're a bit into our experiment and it has had time to warm-up. Let's say we're at $4\pi$ hours: there's enough bacteria that reproduction is looking more and more continuous. Intuitively, the scientists' best model for the number of bacteria at this point is $3 \times 2^{4\pi} = 18,196$.

Illustrating $a^b \times a^c = a^{b+c}$

Let's say we are $i$ hours into the experiment, at which point we have $3 \times 2^{i}$ bacteria. Let's say we wait an additional $j$ hours, at which point we have $3 \times 2^{i+j}$ bacteria.

Let's do a little mental trick and assume that hour $i$ was actually the start of the experiment, where we had $3 \times 2^{i}$ bacteria to start with. Now after $j$ hours, we could alternatively say that we have $3 \times 2^{i} \times 2^{j}$ bacteria.

Here we see intuitively that $3 \times 2^{i+j} = 3 \times 2^{i} \times 2^{j}$. Starting with $3$ bacteria and running the experiment for $i+j$ hours results in the same amount as starting the experiment with $3 \times 2^{i}$ bacteria and running it for $j$ hours.

Illustrating why $a^{-b} = \frac{1}{a^b}$

For this we're going to need some penicillin.

enter image description here

This time, let's say $k$ hours into the experiment, the scientists introduce an antibiotic that will stop the bacteria reproducing and will continuously kill half the bacteria every hour.

At $k+n$ hours (let's stick with the idea that $n<k$) there will be $\frac{1}{2^n}$ times the number of bacteria that there were at the peak of $k$ hours. In other words, there will be $3 \times 2^{k} \times \frac{1}{2^n}$ bacteria at $k+n$ hours.

But, for example, doubling the bacteria for 5 hours and halving for 2 hours is equivalent to just doubling the bacteria for 3 hours. Doubling for $k$ hours and halving for $n$ hours is the same as doubling for $k-n$ hours!

From this, we see that with respect to the number of bacteria at $n+k$ hours, $3 \times 2^{k} \times \frac{1}{2^n}$ is equivalent to saying $3 \times 2^{k-n}$, which is (by the rule of $a^{b+c} = a^b \times a^c$) equivalent to saying $3 \times 2^{k} \times 2^{-n}$.

Hence we see why it's useful to say that $2^{-n} = \frac{1}{2^n}$. And intuitively the same thing would hold if the bacteria were to multiply by a factor of $a$ every hour and then divide by $a$ every hour: hence we see $a^{-b} = \frac{1}{a^b}$.

Illustrating why $a^0 = 1$

Well in the first part, we assumed that we start with 3 bacteria and we discussed how after $h$ hours you will have $3 \times 2^{h}$ bacteria, so if we consider the case with $h = 0$, we see why it makes sense that $2^0$ would be one.

Put another way: as in the previous part with penicillin, let's go back and think about the point $n=k$, where bacteria have been doubling and halving for an equal length of time. As this stage, we would expect the original number of bacteria to be present (since equal hours of doubling and halving cancel out), and our formula gives us $3 \times 2^{k-n} = 3 \times 2^{k-k} = 3 \times 2^0 = 3$ bacteria.

Illustrating why $a^\frac{1}{b} = \sqrt[b]{a}$

Finally, let's forget about the antibiotics and consider again just the phase of continuous reproduction. The rate of increase is $2{\times}$ every hour. But what is our best guess for the ratio of bacteria increase every half hour? Using our formula $2^h$, the factor is $2^{1/2}$ times of course! During reproduction, there will always be $2^{1/2}{\times}$ more bacteria than there were a half hour before.

But what does $2^{\frac{1}{2}}$ mean?

Well we know that there will be a factor of $2$ times more bacteria every subsequent hour, and we know that there will be a factor of $2^{\frac{1}{2}}$ times more bacteria every subsequent half-hour, and we know that two half-hours back-to-back make an hour, so we know that $2^{1/2} \times 2^{1/2} = 2$. From this we see that $2^{\frac{1}{2}} = \sqrt{2}$.

We can apply the same reasoning to a third of an hour, a quarter of an hour, and so forth. This illustrates why $a^\frac{1}{b} = \sqrt[b]{a}$.

Illustrating why $(a^b)^c = a^{bc}$

Okay, let's say our bacteria are multi-cellular.

enter image description here

Let's also say that they are not only doubling in number every hour, but each individual bacterium is doubling in size (doubling in cell-count) every hour too.

So the growth rate of cells in the agar plate is $(2^h)^2$ (since the growth of cells is squared within each bacterium), or equivalently $(2^2)^h$ (since cells are quadrupling every hour), or indeed $2^{2h}$ (since the factor of increase of cells after $j$ hours is the same as the factor of increase of bacteria after $2j$ hours).

See also ...

Image Sources: 1, 2, 3.


We seem to have a real-world understanding of exponentials, for example when we say that a certain quantity is increasing "exponentially".

To make the ingredients of this intuition precise, consider a continuous function $f:\mathbb{R}\to \mathbb{R}$ (think of this as describing a quantity $f(t)$ dependent on time $t$). Then we can define $f$ to be increasing exponentially if for any "interval of time" $s\in \mathbb{R}$, there is some positive rate $r_s>0$ depending on $s$ (think of $r_s$ as the rate of increase of $f$ per $s$ units of time) such that, for any time $t\in \mathbb{R}$, $f(t+s)=r_sf(t)$; that is, if we survey our quantity at regular intervals, it should keep multiplying by a fixed rate that depends only on the length of the interval of time.

Suppose $f$ increases exponentially as in the above definition. Let $r=r_1$ (the rate of increase in $1$ unit of time) and let $a=f(0)$. Then in fact we have $f(t)=a\cdot r^t$. Thus the exponential functions capture this intuitive notion of an exponentially increasing function.

Thus, one way of answering "what is $r^t$?" (for positive rates $r$) is:

Let $f$ be a (nonzero) continuous function increasing exponentially at a rate of $r$ per $1$ unit of time. Then $r^t$ is the rate of increase of $f$ per $t$ units of time.

This boils down to the same thing as kigen's answer, I am simply trying to couch it in our intuition of what it means for a quantity to be increasing exponentially.

This also naturally captures the discrete case, since if $n$ is a whole number, then to find the rate of increase of $f$ per $n$ units of time, we just multiply $r$ with itself $n$ times. What allows us to extend from the discrete case to the continuous case is to think of $r$ not as a quantity, but rather as a rate attached to the auxiliary function $f$.


To be honest, exponentiation loses a nice real-world interpretation pretty quick. What $3^{\sqrt{2}}$ means isn't really easy to say, but what it is is more-or-less just $\displaystyle\lim_{x\to\sqrt{2}} 3^x$, which we can make sense of as we can prove $f:\mathbb{R}^+ \mapsto \mathbb{R}^+,\ x\mapsto 3^x$ is a continuous function. If you were to require some other method of making sense of irrational powers, I'd be pretty stuck; such a function being continuous is the crux of how I make sense of it. Things get weird if we allow a negative base; $(-3)^\sqrt{2}$ is actually imaginary (and will be imaginary for all non-integer exponents) - even with strictly real numbers, you'd have to venture into the complexes to understand them.

Within the complex domain, we have $e^z=e^{\Re(z)}(\cos\theta +i\sin\theta)$, and manipulations with logs and stuff will show why the above number is imaginary, and you might hope that this is as far as you might need to look when trying to figure out some 'meaning', but this isn't the end for the weird behaviour of exponentials. We cannot have functions of the form $z^a$ which are holomorphic on $\mathbb{C}$ when $a$ is not an integer, and so the rule $(zw)^a=z^aw^a$ breaks down when $a$ is not an integer - whereas this rule was a very easy consequence of the 'elementary school' definition! From that, I'm not sure I'll ever see a nice, down-to-earth meaning to assign exponentials for anything other than the naturals..

  • 1
    $\begingroup$ Yep, that was what I was afraid of. I guess intuition is a luxury we can't always have... $\endgroup$
    – baum
    Commented Mar 7, 2014 at 0:57
  • $\begingroup$ It's not so bad in my mind. The definition of exponentiation we get at elementary school is quite obviously a discrete thing; why expect a nice interpretation when things become continuous? We've made a generalisation which takes the previously sensible sentence "multiply $a$ by itself $b$ times" and turns it into something absurd! What could it possibly mean to multiply something $\pi$ times!? $\endgroup$
    – FireGarden
    Commented Mar 7, 2014 at 1:05
  • $\begingroup$ @baum why accept so early? I home someone else can come up with an intuitive explanation some time in future. $\endgroup$
    – user13107
    Commented Mar 7, 2014 at 3:33
  • $\begingroup$ Agreed. Unaccepted this answer; will return in a day or two and reassess. $\endgroup$
    – baum
    Commented Mar 7, 2014 at 4:24

$a^b$ refers to the "multiplicative power" of performing b multiplications by a. This is intuitively obvious with positive integer 'b's, but still holds for fractional and negative values when you put a little brain grease into considering what it means to do 'half a multiplication' or a 'negative multiplication'.

$9^\frac{1}{2}$ is 3 since multiplying by 3 is half the multiplication that multiplying by 9 is. (eg, it's the multiplication which, if done twice, will be equivalent to multiplying by 9). Similarly for $8^\frac{1}{3} = 2$, etc.

$4^{-1}$ is 1/4, since multiplying by 1/4 is 'unmultiplying' once by 4. (eg, it's the operation that gets cancelled out by multiplying once by 4.)

$a^0$ (including $0^0$) is 1, since doing no multiplication to something is the same as multiplying that something by 1.

  • $\begingroup$ But how about irrational numbers? $\endgroup$
    – baum
    Commented Mar 7, 2014 at 15:38
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    $\begingroup$ 0^0 is undefined $\endgroup$
    – evil999man
    Commented Mar 8, 2014 at 9:47
  • $\begingroup$ This explanation even works for complex numbers. Raising something to the i'th power is doing something such that doing it twice is raising it to the -1'th power. $\endgroup$
    – Ypnypn
    Commented Mar 9, 2014 at 2:34

A perspective that I don't see any of the other answers focusing on (although it's implicitly there) is this:

Exponentiation is a correspondence between addition and multiplication.

Think of a number line, with $0$ in the "middle", and tick marks at each integer. Moving a certain distance to the right corresponds to adding a positive number, and adding the same number moves you the same distance, no matter where you are on the line. ("Plus $5$" always means "go five units to the right", whether you start from $8$ or from $8000$.)

Now set up another number line, but on this one, multiplication will take the place of addition. Moving the same distance to the right always means multiplying by the same number, not adding it. (For starters, suppose the number we multiply by to move "one unit" right is some real $a>1$.)

On this number line, it makes sense to put $1$ in the middle — consider how multiplying by $1$ is like adding $0$ — and instead of positive numbers on the right and negatives on the left, we get numbers greater than $1$ on the right, and numbers between $0$ and $1$ on the left.

For example, if $a=2$, then the tick marks to the right of $1$ will be numbered $2,4,8,16,\ldots$.

How do we "translate" from the addition line to the multiplication line?

This, of course, depends on our choice of $a$. Setting the two lines parallel, and reading from the addition line to the multiplication line, we've defined a function $f$, with domain $\left(-\infty,\infty\right)$ and range $\left(0,\infty\right)$. Naturally, we have $f(0)=1$ and $f(1)=a$. That's not the notation we generally use — we denote $f(x)$ as $a^x$, but it's helpful here to write it with more traditional function notation.

This function "turns addition into multiplication" in the following way: If you want to model the addition fact $2+3=5$, you can start at $0$ on the addition-line, then go $2$ units right, then go $3$ units right, and now you're at $5$. If you do those same steps on the multiplication-line, you've modeled $a^2\cdot a^3=a^5$. That's why we have the analogy that exponentiation is "repeated multiplication". It's just like how addition is "repeated $+1$", but instead, it's "repeated $\times a$"

What about the "weird" cases?

This model, much better than repeated addition, accounts for all of the other real exponents, such as negative, rational, and irrational ones:

What's $3^{1/2}$? It's the number on the multiplication-by-$3$-line that occupies the same location where we find $\frac12$ on the addition-line. It must be the case that multiplying by it twice in a row is equivalent to multiplying by $3$ one time. (This makes sense because adding $\frac12$, two times in a row, is equivalent to adding $1$.) Sounds like a square root.

What's $3^\pi$? It's the number on the multiplication-by-$3$-line that occupies the same location where we find $\pi$ on the addition-line. Must be between $27$ and $81$.

How do we really talk about this?

Taking a positive real number $a$, the map $x\mapsto a^x$ is a homomorphism from the additive group of real numbers to the multiplicative group of positive real numbers. If $a=1$, it is the trivial homomorphism, with every element being sent to the identity; for $a\ne 1$, it is in fact an isomorphism. Its inverse map is the base-$a$ logarithm, which changes multiplication back into addition.

This is not a completely rigorous definition, as I haven't talked explicitly about continuity, but that part is implied by the matching number-lines visualization.

  • $\begingroup$ This was really great, and is the second time I've ever, in reading dozens of explanations of exponents, read an answer that was truly intuitive. Well done. $\endgroup$
    – Kyle Baker
    Commented May 20, 2020 at 18:24
  • $\begingroup$ A wonderful explanation, indeed. $\endgroup$ Commented May 6, 2021 at 8:15

Let us consider $a^b$. Then, the concept of exponentiation can be understood as the value of $e^{b\ln a}$, since both this expression and $a^b$ are the same. So, for example, if $b$ is an imaginary number $di$, then $a^b$ denotes a rotation by $d\ln a$ in the complex plane.

However, one can also consider arbitrary exponents $A^B$. This can be interpreted as the (hyper-)volume of a (hyper-)cube in dimension $B$ Euclidean Space $\mathbb{R}^B$ with side length $A$.

How about $2^\pi$, like your example? Assuming you know knowledge about fractals, I believe that $A^C$, with $C\in\overline{\mathbb{Q}}\left(\mbox{ and maybe }C\in\overline{\mathbb{Q}}\cup\mathbb{Q}\right)$ can be interpreted as the volume of a cube in fractal-dimensional Euclidean space $\mathbb{R}^B$ (if such a thing exists - I don't know) with side length $A$. As is probably visible, I am a geometer, and I may not have explained what others might know better. However, I have tried my best to explain my intuition of the meaning of $A^B$.

  • $\begingroup$ I had thought about the hyper-dimensional bodies idea. While that doesn't completely sit well, I suppose that may be the only way of explaining it. $\endgroup$
    – baum
    Commented Mar 7, 2014 at 0:29
  • $\begingroup$ @baum As I've said, there may be other ways that neither of us may have thought of. However, according to me, I believe that the hyper-dimensional bodies idea is the most intuitive. $\endgroup$
    – user122283
    Commented Mar 7, 2014 at 0:34

Exponentiation can be defined by parts:

This is a very roughy way to describe it: When $n$ is a natural number, we can define exponentiation recursively setting $x^{0}=1$ and $x^{n+1}=x \cdot x^n$. In case that $x\not =0$ we can extent this definition for negative integers setting $x^{-n}=1/x^n$ when $n\in \mathbb{Z}^{>0}$. Continue in this way, we define the $n$-th root, $x^{1/n}: = \sup \{y\in \mathbb{R}: y^n<x, \text{ and } y\ge 0\}$ notice that here we restrict our definition to positive real numbers and we use the least upper bound property of the real numbers (the problem of define $n$-th roots for negative real can be addressed when one define complex numbers). Again we extent our definition of exponentiation when $q \in \mathbb{Q}$, $q= m/n$, where $m\in \mathbb{Z}$ and $n\in \mathbb{N}\backslash \{0\}$ by the formula $x^{q}: = (x^{1/n})^m$. In each case we can show that this definition capture all the properties which we would wish that exponentiation had, and also that is well-define (this is particular important because can be a lot of "equivalent" rational numbers).

Having define exponentiation for rational, we can extent one more time this definition for real numbers in general. One very interesting way to interpret $x^r$ when $r\in \mathbb{R}$ and you have all the machinery of limits is by sequences (this is my favorite definition). If $(q_n)$ is a sequence of rational number which converges to $r$, we define $x^r:= \lim_{n\to \infty} x^{q_n}$, of course we have to show that the sequence $(x^{q_n})_{n=0}^\infty$ converges, which is a little tricky (we have to show that is a Cauchy sequence) and is well-define, because there can be multiple choice of sequences which converges to the real number $q$ and they all have to give us the same limit, in ordered that the definition makes sense, fortunately this happens; in particular, if $q_n \to r$ and $q'_n \to r$, both $(x^{q_n})$ and $(x^{q'_n})$ converges to the same value. Also you can check that this definition is consistent with the old one, for example if $r$ is the rational $q$, the sequence $(q)$ trivially converges to $q$ and so $(x^q) \to x^q$. There are others way to define real exponentiation, of course, but in my humble opinion this is a very natural definition because a lot of standard operation in the real numbers are define in a similar way (using limits).

Anyway, I'm a newbie in math but in my opinion thinking about what "really means" real exponentiation can lead to a lot of unnecessary philosophical anguish, similar to thinking about what means the real numbers or the complex. The important thing to know is what properties they have and what we can do with them, no matter whether the exponentiation for natural number represent something in particular in the "real-world", for integers other thing, etc., this is pointless, at least from the mathematical point of view. Because thinking in a specific real world model works well until one is forced to move to another system, in this specific case, a number system like rational, irrational, reals, complex, etc and also in mathematics there is nothing like a good model as long as it does all the right things (all the properties that we would wish exponentiation has) that's good enough to do math. That's the reason for which you can see alternative definitions, but all of them have the same properties so all equally important.


Here is a partial answer that I posted a long time ago: What does $2^x$ really mean when $x$ is not an integer?

In comments above I said I do not believe that power series are the only way to explain exponentials of complex numbers, and that I think there are more intuitive ways. Somone asked for specifics.

I first heard complex multiplication explained as rotating and dilating, so that , for example, multiplying by $i$ (which was called $j$ when I learned this) means rotating $90^\circ$ counterclockwise. Clearly this generalizes multiplication of real numbers, sincec that involves a dilation, and a rotation of either $0^\circ$ or $180^\circ$.

Viewed that way, the base $i$ exponential is obviously about rotating as the argument to the exponential function changes: $i^n$ goes around in a circle as $n$ changes. It's hardly a great leap to imagine $n$ as a non-integer and a smooth motion around the circle.

Now look at $e^z=e^{x+iy}$ where $x$ and $y$ are real. At $z=0$, the rate of change of the exponential function is $1$, so an infinitely small change from $0$ to $0+dz$ results in an infinitely small change in the value of the function from $1$ to $1+(1\cdot dz)$. And that starts it moving counterclockwise around the circle as $z$ moves along the imaginary axis. And as long as it's on the cirlce, the rate of change is tangential to the circle, and that clearly menas the rate of change is a $90^\circ$ counterclockwise rotation of the actual value of the exponential function.


Here is a nice way that I found to make intuitive irrational real powers;


$$\lim_{n\to \infty} (1+r\cdot (\sqrt[n]x-1))^n=x^r$$

Since the numbers $x$ and $r$ are used as they are, and we can choose for $n$ natural numbers as large as we wish, it becomes pretty intuitive. Isn't that?

And here is my proof, or how I came to this theorem:

1) Preparation theorem;

$$\lim_{n\to \infty}n(x^{\frac{1}{n}}-1)=\ln x$$proof;$$e^a=\lim_{n\to \infty}(1+\frac{a}{n})^n \\ e^{\lim_{n\to \infty}n(x^{\frac{1}{n}}-1)}=\lim_{n\to \infty}(1+\frac{n(x^{\frac{1}{n}}-1)}{n})^n \\ e^{\lim_{n\to \infty}n(x^{\frac{1}{n}}-1)}=x\\\lim_{n\to \infty}n(x^{\frac{1}{n}}-1)=\ln x $$

Now to our proof:

$$x^r=e^{\ln x^r}=e^{r\cdot \ln x}=\lim_{n\to \infty}(1+\frac{r\cdot\ln x}{n})^n=\lim_{n\to \infty}(1+\frac{r\cdot n(x^{\frac{1}{n}}-1)}{n})^n\\x^r=\lim_{n\to \infty}(1+r(\sqrt[n]{x}-1))^n$$

Is that good enough?


I think there have been some good answers, at least one really good answer, but I recently had my own epiphany that made exponents 'just make sense' to me, and it's shorter than any of the answers here.

Multiplication is an order higher than addition

$8*1/4$ = the number that, when you add it four times, you get 8

so, $8*1/4 = 2$


$2+2+2+2 = 8$

Exponentiation is an order higher than multiplication

$8^{1/4}$ = the number that, when you multiply it four times, you get 8

so, $8^{1/4}=1.6817[...]$


$1.6817[...]*1.6817*1.6817*1.6817=8$ (if the decimals were truly precise)

(for a cleaner example, note that $8^{1/3}=2$ and $2*2*2=8$.)

I go into more detail in this answer, and include a bibliography of things that helped me or may help others: https://math.stackexchange.com/a/3678743/788921


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