# Exponents Part 2

two

Exponents Part 2

Division

In the previous section we learned that exponents are repeated multiplication, which on its own is not tricky. What makes exponents tricky is determining what is a base and what is not for a given exponent. It is imperative that you really understand the material from the previous section before tackling what’s next. If you did not attempt the practice problems, you need to. Also watch the video that review them.

In this section we are going to see why anything to the power of zero is one and how to handle negative exponents, and why they mean division.

What Happens with Division and Exponents?

Consider the following expression, keeping in mind that the base is arbitrary, could be any number (except zero, which will be explained soon).

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3\cdot 3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 3. Note that 3 is just 31.

$\frac{{3}^{5}}{{3}^{1}}$

If we write this out to seek a pattern that we can use for a short-cut, we see the following:

$\frac{{3}^{5}}{{3}^{1}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3}$

If you recall how we explored reducing Algebraic Fractions, the order of division and multiplication can be rearranged, provided the division is written as multiplication of the reciprocal. That is how division is written here.

$\frac{{3}^{5}}{{3}^{1}}=\frac{3}{3}\cdot \frac{3\cdot 3\cdot 3\cdot 3}{1}$

And of course 3/3 is 1, so this reduces to:

$\frac{{3}^{5}}{{3}^{1}}=3\cdot 3\cdot 3\cdot 3={3}^{4}$

The short-cut is:

$\frac{{3}^{5}}{{3}^{1}}={3}^{5-1}={3}^{4}$

That is, if the bases are the same you can reduce. Reducing eliminates one of the bases that is being multiplied by itself from both the numerator and the denominator. A general form of the third short-cut is here:

Short-Cut 3: $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$

This might seem like a worthless observation, but this will help articulate the very issue that is going to cause trouble with exponents and division.

$\frac{{3}^{5}}{{3}^{1}}={3}^{5}÷{3}^{1}$ .

But that is different than

${3}^{1}÷{3}^{5}$

The expression above is the same as

$\frac{{3}^{1}}{{3}^{5}}$

This comes into play because

$\frac{{3}^{1}}{{3}^{5}}={3}^{1-5}$,

and 1 $–$ 5 = -4.

Negative Exponents?

In one sense, negative means opposite. Exponents mean multiplication, so a negative exponent is repeated division. This is absolutely true, but sometimes difficult to write out. Division is not as easy to write as multiplication.

Consider that 3-4 is 1 divided by 3, four times. 1 ÷ 3 ÷ 3 ÷ 3 ÷ 3. But if we rewrite each of those ÷ 3 as multiplication by the reciprocal (1/3), it’s must cleaner and what happens with a negative exponent is easier to see.

$1÷3÷3÷3÷3\to 1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}$

This is classically repeated multiplication. While one times itself any number of times is still one, let’s go ahead and write it out this time.

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot {\left(\frac{1}{3}\right)}^{4}$

This could also be written:

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot \frac{{1}^{4}}{{3}^{4}}$

The second expression is easier, but both are shown here to make sure you see they are the same.

Since 1 times 14 is just one, we can simplify this further to:

$1\cdot \frac{{1}^{4}}{{3}^{4}}=\frac{1}{{3}^{4}}.$

Negative exponents are repeated division. Since division is hard to write and manipulate, we will write negative exponents as multiplication of the reciprocal. In fact, if instructions say to simplify, you cannot have a negative exponent in your final answer. You must rewrite it as multiplication of the reciprocal. Sometimes that can get ugly. Consider the following:

$\frac{b}{{a}^{-5}}$

To keep this clean, let us consider separating this single fraction as the product of two rational expressions.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{1}{{a}^{-5}}$

The b is not a problem here, but the other rational expression is problematic. We need to multiply by the reciprocal of $\frac{1}{{a}^{-5}}$, which is just a5.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{{a}^{5}}{1}={a}^{5}b$.

This can also be considered a complex fraction, the likes of which we will see very soon. Let’s see how that works.

$\frac{b}{{a}^{-5}}$

Note: ${a}^{-5}=\frac{1}{{a}^{5}}$

Substituting this we get:

$\frac{b}{\frac{1}{{a}^{5}}}$

This is b divided by 1/a5.

$b÷\frac{1}{{a}^{5}}$

Let’s multiply by the reciprocal:

$b\cdot {a}^{5}$

Now we will rewrite it in alphabetical order (a good habit, for sure).

${a}^{5}b$

Let us consider one more example before we make our fourth short-cut. With this example we could actually apply our second short-cut, but it will not offer much insight into how these exponents work with division.

This is the trickiest of all of the ways in which exponents are manipulated, so it is worth the extra exploration.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}$

As you see we have four separate bases. In order to simplify this expression we need one of each base (2, x, y, z), and all positive exponents. So let’s separate this into the product of four rational expressions, then simplify each.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}\to \frac{2}{{2}^{-2}}\cdot \frac{{x}^{-2}}{x}\cdot \frac{{y}^{-5}}{{y}^{3}}\cdot \frac{z}{{z}^{-5}}$

The base of two first:

$\frac{2}{{2}^{-2}}\to 2÷{2}^{-2}$

We wrote it as division. What we will see is dividing is multiplication by the reciprocal, and then the negative exponent is also dividing, which is multiplication by the reciprocal. The reciprocal of the reciprocal is just the original. But watch what happens with the sign of the exponent.

First we will rewrite the negative exponent as repeated division.

$2÷\frac{1}{{2}^{2}}$

Now we will rewrite division as multiplication by the reciprocal.

$2\cdot {2}^{2}={2}^{3}$

Keep in mind, this is the same as 23/1.

We will offer similar treatment to the other bases.

Consider first $\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Negative exponents are division, so:

$\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Notice the x that is already dividing (in the denominator) does not change. It has a positive exponent, which means it is already written as division.

$\frac{{x}^{-2}}{1}\cdot \frac{1}{x}\to \frac{1}{{x}^{2}}\cdot \frac{1}{x}=\frac{1}{{x}^{3}}$

This is exactly how simplifying the y and z will operation.

$\frac{{2}^{3}}{1}\cdot \frac{1}{{x}^{2}\cdot x}\cdot \frac{1}{{y}^{5}\cdot {y}^{3}}\cdot \frac{z\cdot {z}^{5}}{1}$

Putting it all together:

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}=\frac{{2}^{3}{z}^{6}}{{x}^{3}{y}^{8}}$.

Short-Cut 4: Negative exponents are division, so they need to be rewritten as multiplication by writing the reciprocal and changing the sign of the exponent. The last common question is what happens to the negative sign for the reciprocal? What happens to the division sign here: $3÷5=3×\frac{1}{5}$. When you rewrite division you are writing it as multiplication. Positive exponents are repeated multiplication.

${a}^{-m}=\frac{1}{{a}^{m}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{{a}^{-m}}={a}^{m}$

This is the second to last thing we need to learn about exponents. However, a lot of practice is required to master them fully.

To see why anything to the power of zero is one, let’s consider:

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3"#x22C5;3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 35.

$\frac{{3}^{5}}{{3}^{5}}$

Without using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=1$

Using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}={3}^{5-5}$

Five minutes five is zero:

${3}^{5-5}={3}^{0}$

Then 30 = 1.

Τhe 3 was an arbitrary base. This would work with any number except zero. You cannot divide by zero, it does not give us a number.

The beautiful thing about this is that no matter how ugly the base is, if the exponent is zero, the answer is just one. No need to simplify or perform calculation.

${\left(\frac{{3}^{2x-1}\cdot {e}^{\pi i}}{\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}}\right)}^{0}=1$

Let’s take a quick look at all of our rules so far.

 Short-Cut Example ${a}^{m}\cdot {a}^{n}={a}^{m+n}$ ${5}^{8}\cdot 5={5}^{8+1}={5}^{9}$ ${\left({a}^{m}\right)}^{n}={a}^{mn}$ ${\left({7}^{2}\right)}^{5}={7}^{10}$ $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ $\frac{{5}^{7}}{{5}^{2}}={5}^{7-2}={5}^{5}$ ${4}^{-3}=\frac{1}{{4}^{3}}$ ${a}^{0}=1$ 50 = 1

Let’s try some practice problems.

Instructions: Simplify the following.

1. ${\left({2}^{8}\right)}^{1/3}$ 2. $3{x}^{2}\cdot {\left(3{x}^{2}\right)}^{3}$

3. $\frac{5}{{5}^{m}}$ 4. $\frac{{5}^{2}{x}^{-3}{y}^{5}}{{5}^{-3}{x}^{-4}{y}^{-5}}$

5. $7÷7÷7÷7÷7÷7÷7÷7$ 6. $9{x}^{2}y÷9{x}^{2}y$

7. $9{x}^{2}y÷\left(9{x}^{2}y\right)$ 8. ${\left({x}^{2}\cdot 2{x}^{6}\right)}^{2}\cdot {\left({x}^{2}\cdot 2{x}^{6}\right)}^{-2}$

9. ${\left({a}^{m}\right)}^{n}\cdot {a}^{m}$ 10. $\frac{{\left(3{x}^{2}+4\right)}^{2}}{{\left(3{x}^{2}+4\right)}^{3}}$