Often, whether in the sciences or elsewhere, one needs to work with very large values whose factorizations involve some number being multiplied by itself some number of times. Consider the following, for example:

- the speed of light is roughly $300,000,000$ meters per second
- the atomic mass of an element is given in terms of grams per $602,200,000,000,000,000,000,000$ atoms of that element.
- the number of bits contained in a terabyte of data is $8,796,093,022,208$

Doing any calculations with the values shown above can be tedious if they are left in these long forms -- especially if the calculations are done by hand. However, we can take advantage of certain repeated factors of these numbers to write them in a much better form...

Note, in the first two examples the long strings of zeros...

Recall, we can add zeros to the end of any number by multiplying by some number of $10$'s. So, for example, we could turn $3$ into $300,000,000$ by multiplying by ten exactly eight times. Similarly, we could multiply $6.022$ by ten exactly twenty three times to arrive at the number used in the second example.

Of course, there is nothing inherently special about the value $10$, as the third example demonstrates. Here we have a number made solely from a product of two's -- forty three of them, to be exact.

Suppose we adopt the following shorthand: $$x^n = \underbrace{x \cdot x \cdot x \cdots x}_{n\textrm{ times}}$$

As a matter of verbiage, $x^n$ is read as "$x$ raised to the $n^{th}$ power", and the act of raising a value to some power is called **exponentiation**.

We refer to $n$ as the **exponent** on $x$, while $x$ itself is called the **base** of the exponentiation.

As special cases, we call $x^2$ the **square** of $x$, and $x^3$ the **cube** of $x$.

At the moment, let us assume that $n \ge 2$, although we will quickly see that more general values of $n$ can be considered.

Using exponents, we can write all of the numbers in our earlier examples much more succinctly:

- the speed of light is roughly $3 \times 10^8$ meters per second
- the atomic mass of an element is given in terms of grams per $6.022 \times 10^{23}$ atoms of that element.
- the number of bits contained in a terabyte of data is $2^{43}$

Of course, it would be silly to develop this shorthand for writing such products if we needed to go back to the original form everytime we needed to use them in a calculation. As such, we need to develop some rules for what results when powers are combined.

One can combine powers in several ways, of course. We can add two powers together, subtract them, consider their product, or quotient -- we could even raise one power to another. Sums and differences of powers lead us directly into a discussion of polynomials, which we will discuss separately. For now, let us just consider products, quotients, and powers...

Consider the following product:

$$\begin{array}{rcl}

x^3 \cdot x^5 &=& (x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x \cdot x) \\\\

&=& x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\\\

&=& x^8

\end{array}$$

Was there really a need to expand the powers above to find the overall product? Absolutely not! We know $x^3$ contributes three factors of $x$ to the overall product, and $x^5$ contributes five more, so $x^3 \cdot x^5$ must be the product of $3+5=8$ factors of $x$ (that is to say, $x^8$). This suggests the following rule:$\require{cancel}$

$$\boxed{\displaystyle{x^m \cdot x^n = x^{m+n}}}$$

Taking this a bit further, consider a power that is in turn raised to some other power:

$$\begin{array}{rcl}

(x^m)^n &=& \underbrace{x^m \cdot x^m \cdot \cdots \cdot x^m}_{\textrm{n times}}\\\\

&=& x^{m + m + \cdots + m}\\\\

&=& x^{m\,n}

\end{array}$$

which suggests that raising a power to a power should be dealt with by multiplying the exponents:

$$\boxed{\displaystyle{(x^m)^n = x^{m\,n}}}$$

Turning our attention away from products of powers, and focusing now on quotients of powers -- consider the following:

$$\begin{array}{rcl}

\frac{x^5}{x^3} &=& \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x}\\\\

&=& \frac{x \cdot x \cdot \cancel{x \cdot x \cdot x}}{\cancel{x \cdot x \cdot x}}\\\\

&=& x \cdot x\\\\

&=& x^2

\end{array}$$

Here again, is it necessary to expand the powers above to find the overall quotient? Certainly not! Notice, the numerator and denominator share the same base, and the exponent involved with the denominator is less than that associated with the numerator. As such, three of the factors of $x$ in the numerator and denominator will cancel,^{†} leaving a product of $5-3=2$ factors of $x$ (equivalently, $x^2$). Here again, a more general rule is suggested:

$$\boxed{\displaystyle{\frac{x^m}{x^n} = x^{m-n}, \quad \textrm{provided } x \ne 0}}$$

The rule above has an unexpected consequence however. To see this, consider which values of $n$ and $m$ make sense. Notice, even keeping $n,m \ge 2$, one could easily have $m-n \lt 2$. However, If we limit ourselves to defining $x^n$ as the product of $n$ factors of $x$ though, powers like $x^0, x^{-1}, x^{-2},\ldots$ simply don't have any meaning. Truthfully, even $x^1$ is on a shaky ground -- how can you have a "product" with only a single factor?

So, should we shun negative powers completely? Should we disallow the use of this rule for simplifying the quotient of powers if the resulting exponent is negative? Or should we generalize the definition of $x^n$ so that zero and negative exponents actually do have meaning?

We opt for the last option -- and nicely enough, going back to the expanded form for $x^n$ tells us exactly how such negative exponents should be interpreted. Consider the following:

suggested rule

$$\begin{array}{rcl}

\frac{x^3}{x^5} &=& x^{3-5}\\

&=& x^{-2}

\end{array}$$

to the definition

$$\begin{array}{rcl}

\frac{x^3}{x^5} &=& \frac{x \cdot x \cdot x}{x \cdot x \cdot x \cdot x \cdot x}\\\\

&=& \frac{\cancel{x \cdot x \cdot x}}{x \cdot x \cdot \cancel{x \cdot x \cdot x}}\\\\

&=& \frac{1}{x \cdot x}\\\\

&=& \frac{1}{x^2}

\end{array}$$

So to be consistent, we should interpret $x^{-2}$ as $\displaystyle{\frac{1}{x^2}}$

Similar calculations suggest the following definition for dealing with negative exponents:

$$\boxed{\displaystyle{x^{-n} = \frac{1}{x^n} \quad \textrm{provided } x \ne 0}}$$

In a like manner, we might "discover" how we should interpret any value to a zero power through the following calculations (assume $x \neq 0$, so that we avoid divisions by zero):

suggested rule

$$\begin{array}{rcl}

\frac{x^3}{x^3} &=& x^{3-3}\\

&=& x^{0}

\end{array}$$

to the definition

$$\begin{array}{rcl}

\frac{x^3}{x^3} &=& \frac{x \cdot x \cdot x}{x \cdot x \cdot x}\\\\

&=& \frac{\cancel{x \cdot x \cdot x}}{\cancel{x \cdot x \cdot x}}\\\\

&=& 1\\\\

\end{array}$$

Again - for the sake of consistency, we make the following definition:

$$\boxed{\displaystyle{x^0 = 1 \quad \textrm{for any } x \neq 0}}$$

The same trick can be used to suggest the following definition for $x^1$, lest the earlier-mentioned "product of one factor" gave anyone pause. Just consider $x^4 / x^3$, for example...

$$\boxed{\displaystyle{x^1 = x}}$$

Having considered products and quotients of powers, we now turn our attention to the reverse -- powers of products and quotients. Recall how multiplication distributes over addition and subtraction:

$$a(b + c) = ab + ac \quad \quad \textrm{and} \quad \quad a(b - c) = ab - ac$$

In a similar manner, exponentiation "distributes" over multiplication and division. That is to say (presuming the expressions below are defined):

$$\boxed{\displaystyle{(xy)^n = x^n y^n \quad \quad \text{and} \quad \quad \left( \frac{x}{y} \right)^n = \frac{x^n}{y^n}}}$$

These can be easily shown to hold true for integer exponents $n \ge 2$, as evinced below:

$$\begin{array}{rcl}

(xy)^n &=& \underbrace{xy \cdot xy \cdot \cdots \cdot xy}_{\textrm{$n$ factors of $xy$}} \\\\

&=& \underbrace{x \cdot x \cdot \cdots \cdot x}_{\textrm{$n$ factors of $x$}} \cdot \underbrace{y \cdot y \cdot \cdots \cdot y}_{\textrm{$n$ factors of $y$}}\\\\

&=& x^n y^n

\end{array}$$

(xy)^n &=& \underbrace{xy \cdot xy \cdot \cdots \cdot xy}_{\textrm{$n$ factors of $xy$}} \\\\

&=& \underbrace{x \cdot x \cdot \cdots \cdot x}_{\textrm{$n$ factors of $x$}} \cdot \underbrace{y \cdot y \cdot \cdots \cdot y}_{\textrm{$n$ factors of $y$}}\\\\

&=& x^n y^n

\end{array}$$

$$\begin{array}{rcl}

\left( \frac{x}{y} \right)^n &=& \underbrace{\frac{x}{y} \cdot \frac{x}{y} \cdot \cdots \cdot \frac{x}{y}}_{\textrm{$n$ factors of $\frac{x}{y}$}}\\\\

&=& \frac{x \cdot x \cdot \cdots \cdot x}{y \cdot y \cdot \cdots \cdot y}\\\\

&=& \frac{x^n}{y^n}

\end{array}$$

\left( \frac{x}{y} \right)^n &=& \underbrace{\frac{x}{y} \cdot \frac{x}{y} \cdot \cdots \cdot \frac{x}{y}}_{\textrm{$n$ factors of $\frac{x}{y}$}}\\\\

&=& \frac{x \cdot x \cdot \cdots \cdot x}{y \cdot y \cdot \cdots \cdot y}\\\\

&=& \frac{x^n}{y^n}

\end{array}$$

As a useful matter of convention, when simplifying expressions that contain exponents by using the above rules, one should avoid the presence of negative exponents in the final form.