Multiplying Exponents

In this tutorial, we will learn about multiplying exponents. And dividing them.

But first, a quick recap of what exponents are.

Exponents

Exponents (also known as indices or powers) are shorthand for repeated multiplication. The exponent of a number tells us how many copies of it are multiplied together.

For example –

With this in mind, let’s move ahead.

The Two Basic Rules

When it comes to multiplying exponents, there are two simple rules you need to keep in mind.

1. Multiplying Exponents When Bases Are the Same

When you are multiplying exponents with the same base, add the exponents and copy the base.

Let’s see this rule in action.

Example

Simplify.

(a)52⋅5(a) \hspace{0.75em} 5^2 \cdot 5
(b)23⋅22⋅25(b) \hspace{0.75em} 2^3 \cdot 2^2 \cdot 2^5

Solution (a)

We have the same base here (5), so we will add the exponents. Also, remember, when you don’t see an exponent, there’s always a “1” there. So –

52⋅5=52⋅51=5(2+1)=53=125\begin{align*} 5^2 \cdot {\color{Red} 5} &= 5^2 \cdot {\color{Red} 5^1} \\[1em] &= 5^{(2+1)} \\[1em] &= 5^3 \hspace{0.25em} = \hspace{0.25em} 125 \end{align*}

Easy, right?

Solution (b)

Here, we have more than two numbers being multiplied. But that doesn’t make any difference. The rule remains the same. Add the exponents if they have the same base.

23⋅22⋅25=2(3+2+5)=210=1032\begin{align*} 2^3 \cdot 2^2 \cdot 2^5 &= 2^{(3+2+5)} \\[1em] &= 2^{10} \hspace{0.25em} = \hspace{0.25em} 1032 \end{align*}
How It Works

To help you gain a more intuitive understanding of how the rule works, here’s an example showing the intermediate step.

Now, in place of 5, 2, and 3, you can have any number. This would still be true. Hence, the rule.

2. Multiplying When Exponents Are the Same

When multiplying exponents, if you have the same exponent on different bases, multiply the bases and keep the same exponent.

Let’s look at a couple of examples.

Example

Simplify.

(a)23⋅53(a) \hspace{0.75em} 2^3 \cdot 5^3
(b)22⋅62⋅102(b) \hspace{0.75em} 2^2 \cdot 6^2 \cdot 10^2

Solution (a)

In this example, we have the same exponent – 3. So we keep that exponent and multiply the bases together.

23⋅53=(2⋅5)3=103=1000\begin{align*} {\color{Teal} 2} ^ {\color{Red} 3} \cdot {\color{Orchid} 5} ^ {\color{Red} 3} &= ( {\color{Orchid} 2} \cdot {\color{Orchid} 5} )^ {\color{Red} 3} \\[1em] &= 10^ {\color{Red} 3} \\[1em] &= 1000 \end{align*}

That’s it.

Solution (b)

Again, more numbers here but that’s okay. As long as the exponents are the same, we can use our simple rule.

22⋅62⋅102=(2⋅6⋅10)2=1202=14400\begin{align*} 2^2 \cdot 6^2 \cdot 10^2 &= (2 \cdot 6 \cdot 10)^2 \\[1em] &= 120^2 \hspace{0.25em} = \hspace{0.25em} 14400 \end{align*}
How It Works

Here’s a quick example to help you with the intuition for this rule.

Say we are multiplying 23\hspace{0.2em} 2^3 \hspace{0.2em} and 53\hspace{0.2em} 5^3 \hspace{0.2em}.

23⋅53=2⋅2⋅2⋅5⋅5⋅5 {\color{Red} 2^3} \cdot {\color{Teal} 5^3} \hspace{0.25em} = \hspace{0.25em} {\color{Red} 2} \cdot {\color{Red} 2} \cdot {\color{Red} 2} \cdot {\color{Teal} 5} \cdot {\color{Teal} 5} \cdot {\color{Teal} 5}

Now because they had the same exponent, both 2\hspace{0.2em} 2 \hspace{0.2em} and 5\hspace{0.2em} 5 \hspace{0.2em} repeat the same number of times on the right side – 3\hspace{0.2em} 3 \hspace{0.2em} times each. So we can group them into 3\hspace{0.2em} 3 \hspace{0.2em} pairs.

23⋅53=(2⋅5)⋅(2⋅5)⋅(2⋅5) {\color{Red} 2^3} \cdot {\color{Teal} 5^3} \hspace{0.25em} = \hspace{0.25em} ( {\color{Red} 2} \cdot {\color{Teal} 5} ) \cdot ( {\color{Red} 2} \cdot {\color{Teal} 5} ) \cdot ( {\color{Red} 2} \cdot {\color{Teal} 5} )

And we arrive at the expected result.

Again, we could replace 2\hspace{0.2em} 2 \hspace{0.2em}, 5\hspace{0.2em} 5 \hspace{0.2em}, and 3\hspace{0.2em} 3 \hspace{0.2em} with any three numbers and the basic idea would still apply. Hence, the rule.

But What if Both the Bases and Exponents Are different?

If both bases and exponents are different, unfortunately, there are no magic rules like we saw above to simplify them. Generally, we need to work with each exponent individually.

Here’s an example.

Simplify : 23⋅52\hspace{0.2em} \hspace{1em} 2^3 \cdot 5^2 \hspace{0.2em}

Solution

In this example, bases, as well as the exponents, are different. So we can’t use either of the two rules mentioned earlier. We’ll have to simplify the two exponents separately.

23⋅52=8⋅25=200\begin{align*} {\color{Red} 2^3} \cdot {\color{Teal} 5^2} &= {\color{Red} 8} \cdot {\color{Teal} 25} \\[1em] &= 200 \end{align*}

This was a simple example. So we didn’t really miss those rules. But don’t let that make you underestimate their importance.

Some Special Cases

Alright, now let’s use the two rules we have learned so far and apply them to some special cases you can come across when multiplying exponents.

Negative Exponents

When adding negative exponents, make sure you add them with their signs.

Sounds confusing? Don’t worry. The following examples should help remove any doubts you have.

Example

Simplify.

(a)3−4⋅32(a) \hspace{0.75em} 3^{-4} \cdot 3^2
(b)54⋅5−2⋅102(b) \hspace{0.75em} 5^4 \cdot 5^{-2} \cdot 10^2

Solution (a)

Here the bases are the same, so we add the exponents. But we add -4 with its negative sign. So effectively, it gets subtracted.

34×32=3(−4+2)=3−2\begin{align*} 3^4 \times 3^2 &= 3^{(-4 + 2)} \\[1em] &= 3^{-2} \end{align*}

Also, as I explain in the tutorial on negative exponents, you can change the sign of the exponent by taking the reciprocal (interchanging the top and bottom numbers) of the base.

So here, we move 3 to the bottom and change its exponent from -2 to 2.

3−2=132=19\begin{align*} 3^{-2} &= \frac{1}{3^2} \\[1.3em] &= \frac{1}{9} \end{align*}

Solution (b)

Nothing too different here. We add the exponents with their signs.

54×5−2=5(4+(−2))=5(4−2)=52=25\begin{align*} 5^4 \times 5^{-2} &= 5^{(4 + (-2))} \\[1em] &= 5^{(4 - 2)} \\[1em] &= 5^2 \hspace{0.25em} = \hspace{0.25em} 25 \end{align*}
Example

Simplify.

(a)5−3⋅2−3(a) \hspace{0.75em} 5^{-3} \cdot 2^{-3}
(b)10−2⋅2−2(b) \hspace{0.75em} 10^{-2} \cdot 2^{-2}

Solution (a)

This time, we have the same (negative) exponent. So we copy it over without doing anything to it. So positive or negative, doesn’t matter.

5−3×2−3=(5×2)−3=10−3=1103=11000\begin{align*} 5^{-3} \times 2^{-3} &= (5 \times 2)^{-3} \\[1.3em] &= 10^{-3} \\[1.3em] &= \frac{1}{10^3} \hspace{0.25em} = \hspace{0.25em} \frac{1}{1000} \end{align*}

Again, by taking the reciprocal, we turned the exponent from negative to positive.

Solution (b)

Same story.

10−2×2−2=(10×2)−2=20−2=1202=1400\begin{align*} 10^{-2} \times 2^{-2} &= (10 \times 2)^{-2} \\[1.3em] &= 20^{-2} \\[1.3em] &= \frac{1}{20^2} \hspace{0.25em} = \hspace{0.25em} \frac{1}{400} \end{align*}

Fractional Exponents

Whether the exponents are whole or fractional, it’s all the same as far as the two rules are concerned.

Simplify.

Example

Simplify.

(a)32/3×52/3(a) \hspace{0.75em} 3^{2/3} \times 5^{2/3}
(b)61/5×62/5(b) \hspace{0.75em} 6^{1/5} \times 6^{2/5}

Solution (a)

Here, we have the same exponent. So we multiply the bases and copy the exponent.

32/3×52/3=(3×5)2/3=152/3\begin{align*} 3^{2/3} \times 5^{2/3} &= (3 \times 5)^{2/3} \\[1em] &= 15^{2/3} \end{align*}

Solution (b)

This time, we have the same base. So we add the exponents and retain the base.

61/5×62/5=6(1/5+2/5)=63/5\begin{align*} 6^{1/5} \times 6^{2/5} &= 6^{(1/5 + 2/5)} \\[1em] &= 6^{3/5} \end{align*}

So, apart from the fact that we needed to add fractions, there was nothing different.

Variables

So far, all of the examples had numbers. But what if we have variables with exponents on them?

When multiplying variables with exponents, the key thing is – treat each variable as a separate item and add exponents only for identical (same) variables.

Example

Simplify.

(a)x2y⋅xy3(a) \hspace{0.75em} x^2y \cdot xy^3
(b)pqr2⋅p2q3r4(b) \hspace{0.75em} pqr^2 \cdot p^2q^3r^4

Solution (a)

In this example, we have two different variables – x and y. So we make two groups – one for x and another for y. And then simplify each group separately.

x2y⋅xy3=x2⋅x⋅y⋅y3=x(2+1)⋅y(1+3)=x3⋅y4=x3y4\begin{align*} {\color{Red} x^2} {\color{Teal} y} \cdot {\color{Red} x} {\color{Teal} y^3} &= {\color{Red} x^2} \cdot {\color{Red} x} \cdot {\color{Teal} y} \cdot {\color{Teal} y^3} \\[1em] &= {\color{Red} x^{(2 + 1)}} \cdot {\color{Teal} y^{(1 + 3)}} \\[1em] &= x^3 \cdot y^4 \hspace{0.25em} = \hspace{0.25em} x^3y^4 \end{align*}

And that’s it. We can’t simplify it any further.

Solution (b)

Here we have three variables, so we make three groups. Everything else remains the same.

pqr2⋅p2q3r4=p⋅p2⋅q⋅q3⋅r2⋅r4=p(1+2)⋅q(1+3)⋅r(2+4)=p3q4r6\begin{align*} pqr^2 \cdot p^2q^3r^4 &= p \cdot p^2 \cdot q \cdot q^3 \cdot r^2 \cdot r^4 \\[1em] &= p^{(1 + 2)} \cdot q^{(1 + 3)} \cdot r^{(2 + 4)} \\[1em] &= p^3q^4r^6 \end{align*}

Dividing Exponents

As you will see, dividing exponents is very similar to multiplying exponents, except for one important difference.

Let’s look at the two cases – same base and same exponent – one by one, with examples.

Same Base

When dividing exponents with the same base, we subtract the exponents instead of adding them (as we do in multiplication).

Example

Simplify.

(a)3533(a) \hspace{0.75em} \frac{3^5}{3^3}
(b)p3q2rpqr2(b) \hspace{0.75em} \frac{p^3q^2r}{pqr^2}

Solution (a)

We have the same base (3) here. Had it been multiplication, we would add the exponents. But since this is division, we subtract the exponent of the divisor (the second or bottom one).

3533=3(5−3)=32=9\begin{align*} \frac{3^5}{3^3} &= 3^{(5 - 3)} \\[1.3em] &= 3^2 \hspace{0.25em} = \hspace{0.25em} 9 \end{align*}

Done!

Solution (b)

In this example, we have variables. So we’ll follow the same strategy we used when multiplying exponents. We’ll group identical variables together and process each group individually.

p3q2rpqr2=p3p⋅q2p⋅rr2=p(3−1)⋅q(2−1)⋅r(1−2)=p2⋅q⋅r−1=p2qr\begin{align*} \frac{ {\color{Red} p^3} {\color{Teal} q^2} {\color{Orchid} r} }{ {\color{Red} p} {\color{Teal} q} {\color{Orchid} r^2} } &= {\color{Red} \frac{p^3}{p} } \cdot {\color{Teal} \frac{q^2}{p} } \cdot {\color{Orchid} \frac{r}{r^2} } \\[1.3em] &= {\color{Red} p^{(3-1)}} \cdot {\color{Teal} q^{(2-1)}} \cdot {\color{Orchid} r^{(1 - 2)}} \\[1.3em] &= {\color{Red} p^2} \cdot {\color{Teal} q} \cdot {\color{Orchid} r^{-1}} \hspace{0.25em} = \hspace{0.25em} \frac{ {\color{Red} p^2} {\color{Teal} q} }{ {\color{Orchid} r} } \end{align*}

And, at the risk of sounding repetitive – in the last step, we took the reciprocal of r to turn its negative exponent into a positive one.

Same Exponent

When exponents are the same, we do the division on the bases and copy the exponent.

Example

Simplify.

(a)4585(a) \hspace{0.75em} \frac{4^5}{8^5}
(b)(xy)3(yz)3(b) \hspace{0.75em} \frac{(xy)^3}{(yz)^3}

Solution (a)

Alright, here we have the same exponent. So we do the division with the bases and copy the exponent.

4585=(48)5=(12)5=1525=132\begin{align*} \frac{4^5}{8^5} &= \left ( \frac{4}{8} \right )^5 \\[1.3em] &= \left ( \frac{1}{2} \right )^5 \\[1.3em] &= \frac{1^5}{2^5} \hspace{0.25em} = \hspace{0.25em} \frac{1}{32} \end{align*}

Solution (b)

Nothing too different here. Except that we have variables to work with.

(xy)3(yz)3=(xyyz)3=(xz)3=x3z3\begin{align*} \frac{(xy)^3}{(yz)^3} &= \left ( \frac{x \cancel{y}}{\cancel{y} z} \right )^3 \\[1.3em] &= \left ( \frac{x}{z} \right )^3 \hspace{0.25em} = \hspace{0.25em} \frac{x^3}{z^3} \end{align*}

Well, that brings us to the end of this tutorial on multiplying exponents. Make sure you practice a lot; that’s the key to being good at math. Until next time.