Easy Exponents

A cheat sheet for exponents

In the beginning, there was addition. And it begat multiplication — serial addition. And it produced subtraction (addition in reverse) and division (upside down multiplication). Lastly among the lowlier operations came exponentiation — serial multiplication. But ultimately, it’s all just addition.

We learned basic exponentiation in school: x³ means x times x times x (serial multiplication). Great as far as it goes, but what about exponents like this:

\(\displaystyle{x}^{-3}=\,?,\;\;\;\;{x}^{\frac{1}{2}}=\,?\)

How does one serial multiply something minus three times? Or one-half times?

It turns out that it’s pretty simple, and it all boils down the basic fact we learned in school: the exponent says how many times to serial multiply the base number (which we’re calling x here). And it works for those weird cases, too.

The Axiom

It begins with this basic axiom:

\(\displaystyle{x}^{n}={x}_{1}\,\times\,{x}_{2}\,\times\,{x}_{3}\,\times\ldots\times\,{x}_{n}\)

Which just describes mathematically what we’ve said above about serial multiplication. A number to the nth power is multiplied times itself n times.

As an aside, the fancy way to say it mathematically is:

\(\displaystyle{x}^{n}=\prod^{n}_{i=1}{x}_{i}\)

Which says the same thing as the equation above but more succinctly (albeit perhaps also more mysteriously to the uninitiated).

The First Theorem

From the axiom we can easily derive the first theorem:

\(\displaystyle{x}^{1}={x}\)

Any number (any x) to the power of one is just that number because there is just one instance of it in the serial multiplication chain.

[In a very real sense, under the hood, any ordinary multiplication begins at 1. For example, three times four really means 1 × 3 × 4. So, x¹ really means 1 × x, and x² really means 1 × x × x, and so on. This will become more relevant below.]

The Second Theorem

We can also derive the second theorem; one so important it’s often presented as an axiom — as the axiom of exponents:

\(\displaystyle{x}^{a+b}={x}^{a}\times{x}^{b}\)

Note that a+b=n, so it is also true that:

\(\displaystyle{x}^{a+b}={x}^{n}\)

So, we need to do serial multiplication n (= a+b) times:

\(\displaystyle{x}^{a+b}=({x}_{1}\!\times\!{x}_{2}\,\ldots\times\!{x}_{a})\times({x}_{1}\!\times\!{x}_{2}\,\ldots\times\!{x}_{b})={x}^{n}\)

By considering the two parts in parenthesis, we see that parts of the exponent (a and b here) that add can be separated into multiple terms that multiply. As you’ll see, this is super useful, and we’ll come back to it later (because it’s what underlies logarithms and slide rules).

More importantly at the moment, this theorem lets us prove a number of other important theorems.

The Third Theorem

The first of those, the third theorem, is:

\(\displaystyle{x}^{0}={1}\)

Any number (any x) to the power of zero is one.

To see why, note that:

\(\displaystyle{x}^{1+0}={x}^{1}={x}\)

And therefore, using the second theorem:

\(\displaystyle{x}^{1+0}={x}^{1}\times{x}^{0}={x}\)

If we drop the first term and divide the last two terms by x¹ we have:

\(\displaystyle\frac{{x}^{1}\times{x}^{0}}{{x}^{1}} =\frac{x}{{x}^{1}}\)

The x¹ terms on the left cancel out leaving us with:

\(\displaystyle{x}^{0}=\frac{x}{x^1}=\frac{x}{x}={1}\)

So, any x to the zero power is just one. (This gets confused with 0⁰, which is considered undefined but which many calculators and libraries treat as one or zero depending on how the designer saw things.)

The Fourth Theorem

A somewhat similar proof gives us the fourth theorem:

\(\displaystyle{x}^{-n}=\frac{1}{x^n}\)

Negative exponents give us the inverse (the one-over) of the exponentiated value (with a positive exponent).

To see why, note that:

\(\displaystyle{x}^{(n-n)}={x}^{0}=1\)

We can re-express the first term:

\(\displaystyle{x}^{n-n}={x}^{n+(-n)}\)

Which makes it clear we can apply the second theorem:

\(\displaystyle{x}^{n+(-n)}={x}^{n}\times{x}^{-n}=1\)

Now, drop the first term and divide the last two by x^n:

\(\displaystyle\frac{{x}^{n}\times{x}^{-n}}{{x}^{n}} =\frac{1}{{x}^{n}}\)

The terms on the left cancel, leaving us with the fourth theorem.

The Fifth Theorem

Fractional exponents might seem impossible, but they follow from the fifth theorem:

\(\displaystyle{x}^{\frac{1}{n}}=\sqrt[n]{x}\)

The simplest version involves the fraction 1/2:

\(\displaystyle{x}^{\frac{1}{2}+\frac{1}{2}}={x}^{1}={x}\)

We can apply the second theorem to get:

\(\displaystyle{x}^{\frac{1}{2}+\frac{1}{2}}={x}^{\frac{1}{2}}\times{x}^{\frac{1}{2}}=\left({x}^{\frac{1}{2}}\right)^{2}={x}\)

And if we keep only the last two terms and take their square roots:

\(\displaystyle{x}^{\frac{1}{2}}=\sqrt{x}\)

Similarly, for any n:

\(\displaystyle{x}^{\frac{1}{n}\times{n}}={x}^{\frac{n}{n}}={x}^{1}={x}\)

And so:

\(\displaystyle{x}^{\frac{1}{n}+\cdots+\frac{1}{n}}={x}^{\frac{1}{n}}\times\cdots\times{x}^{\frac{1}{n}}=\left({x}^{\frac{1}{n}}\right)^{n}={x}\)

And if we keep the last two terms and take their nth root:

\(\displaystyle{x}^{\frac{1}{n}}=\sqrt[n]{x}\)

So, simple fractions are just roots. But what about fractions such as:

\(\displaystyle{x}^{\frac{3}{4}}=\;?\)

I’ll come back to that in a jiffy. Because first we need…

The Sixth Theorem

Nested exponents fall under the aegis of the sixth theorem:

\(\displaystyle\left({x}^{a}\right)^{b}={x}^{ab}={x}^{ba}=\left({x}^{b}\right)^{a}\)

To see why, note that:

\(\displaystyle\left({x}^{a}\right)^{b}=({x}^{a})_{1}\times({x}^{a})_{2}\times\ldots\times({x}^{a})_{b}\)

And, of course, the inner terms expand to:

\(\displaystyle{x}^{a}={x}_{1}\times{x}_{2}\times\ldots\times{x}_{a}\)

So, all together there are a × b occurrences of x to serial multiply.

Note however:

\(\displaystyle\left({x}^{a}\right)^{b}\ne{x}^{\left({a}^{b}\right)}\)

Note also that, generally:

\(\displaystyle\begin{matrix}{x}^{a}\end{matrix}^{b}=\left({x}^{a}\right)^{b}\)

When there are no parentheses, the precedence moves left-to-right.

A Jiffy Later…

This theorem allows us to answer the question posed above about fractional exponents:

\(\displaystyle{x}^{\frac{3}{4}}={x}^{{3}\times\frac{1}{4}}=\left({x}^{3}\right)^{\frac{1}{4}}=\sqrt[4]{x^{3}}\)

And note they can be nested in either order:

\(\displaystyle{x}^{\frac{3}{4}}={x}^{\frac{1}{4}\times{3}}=\left({x}^{\frac{1}{4}}\right)^{3}=\left(\sqrt[4]{x}\right)^{3}\)

The result is the same either way.

For that matter, we can also break it down this:

\(\displaystyle{x}^{\frac{3}{4}}={x}^{\frac{2}{4}+\frac{1}{4}}={x}^{\frac{1}{2}+\frac{1}{4}}=\sqrt[2]{x}\times\sqrt[4]{x}\)

Which gives the same result.

That’s not quite the end of the story, but it’s enough for now and should get you through most exponential situations.

Logarithms

The log function is the opposite of exponentiation. That is:

\(\displaystyle\text{given:}\;\;{x}^{a}={V},\;\;\text{then:}\;\log_{x}(V)={a}\)

Where x is the base number (typically 2, 10, or e) and a is the exponent. The result of x to the power of a is the value V. The base x log of value V is a, the exponent.

Which gives the following identity:

\(\displaystyle{x}^{a}={V}={x}^{\log_{x}(V)}\)

Because:

\(\displaystyle\log_{x}(V)=a\)

We see the power of logarithms when we combine the above with the second theorem of exponents:

\(\displaystyle{A}\times{B}={x}^{\log_{x}(A)}\times{x}^{\log_{x}(B)}={x}^{\log_{x}(A)+\log_{x}(B)}\)

Which lets us multiply two numbers together (an expensive operation, especially with big numbers) by adding their logarithms. This is exactly what slide rules do. They use logarithmic scales sliding against each other (which is addition).

Take note of a couple of important log identities:

\(\displaystyle\log_{x}(1)=0\;\;\;\text{because}\;\;\;{x}^{0}={1}\)

And:

\(\displaystyle\log_{x}(x)=1\;\;\;\text{because}\;\;\;{x}^{1}={x}\)

Important: The log function is not valid for inputs zero or less because:

\(\displaystyle{0}\lt{x}^{a}\)

For all positive values of x. Of course, -1³=-1, so this doesn’t hold for negative bases, but there’s no negative bases for logarithms, so you can only take the log of values greater than zero.

As a last note, you can convert between exponential bases with the log function:

\(\displaystyle{A}^{x}={B}^{{x}\times\log_{B}(A)}\)

So, for example:

\(\displaystyle{10}^{24}={2}^{{24}\times\log_{2}(10)}={2}^{79.726\ldots}\)

And on that note, I think that’s enough for now!

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