For instance, we can think of exponentials as solutions to differential equations. Are there any other interesting ways of describing exponents without iteration of multiplication?
4 Answers
Using the exponential function (with different characterisations) $e^x$ and the natural logarithm:
For every $a, b \in \mathbb R$, we can write $a = e^{ln(a)}$ (suppose for now $a>0$) which implies $$a^b = (e^{ln(a)})^b = e^{b\cdot ln(a)}$$ See these great answers for more detail.
In fact, this allows to expand the concept of exponentials to the complex plane:
Given $z, w \in \mathbb C$, we define $z^w = e^{w\cdot ln(z)}$ (where $ln(z)$ is the complex logarithm), and use the fact that for $w = a + i\cdot b$:
$$e^w = e^{a+ib}=e^a\cdot e^{ib} = e^a(cos(b)+i\cdot sin(b))$$ since $exp(x+y)= exp(x)\cdot exp(y)$ and $e^{ib} = cos(b)+i\cdot sin(b)$.
However, note that this is a multivalued function unless a branch of the complex logarithm is chosen (otherwise $(e^{ln(z)})^w \neq e^{w\cdot ln(z)}$).
If a function's growth rate is proportional to its value then it's an exponential function.
i.e. $b^x$ is growing at the rate $c_b*b^x$ for some constant $c_b$. $b^x$ is the one function where that can be true.
Which allows for the "how can you multiply something by itself 1/2 a time to get $b^{1/2}$" question to be "how much do you have at 1/2 if the instantaneous grate of growth always proportional to the current ammount?" Of course, that's a pretty hard question to answer without circular reasoning, but it does make noninteger and even non rational exponents meaningful.
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I like this different answer. $\color{white}{+1}$ – Simply Beautiful Art Sep 27 '16 at 20:51
From the category theory point of view, you can adress a lot of exponentials.
For example, in the category of Sets, an exponent of two sets is the set of functions from one to another
( A hint to see why it makes sense: how many functions exists from a set $A$ to a set $B$?
Answer:$B^{A}$ where $X$ is the cardnality of a set $X$
)
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If you go one step beyond multiplication...
$$a^b=\underbrace{a\times a\times a\times\cdots\times a}_b$$
$$a\times b=\underbrace{a+a+a+\dots+a}_b$$
And combine these to turn exponentiation into addition.
$$\color{white}{\text{Way too many additions, so I left it out.}}$$
Take these farther steps, and you'll create the hyperoperation sequence.
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