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Section2.2Properties of Exponents

Consider the product \(x^2x^4\text{.}\) The expression \(x^2\) contains \(2\) factors of \(x\) while the factor of \(x^4\) contains \(4\) factors of \(x\text{,}\) so altogether the expression \(x^2x^4\) contains \(6\) factors of \(x\text{.}\) Note that \(6\) comes from summing \(2\) and \(4\text{.}\) Writing this out using exponent notation we have:

\begin{align*} x^2x^4\amp=x^{2+4}\\ \amp=x^6 \end{align*}

From the above example we can infer the following rule:

The Product Rule for Exponents: \(x^mx^n=x^{m+n}\)

Note that this rule only applies to exponential expressions with a common base (the thing being raised to a power). For example,

\begin{align*} t^6t^{19}\amp=t^{6+19}\\ \amp=t^{25} \end{align*}

On the other hand, there is no way to combine the exponents in the expression \(x^6y^{19}\text{.}\)

When simplifying products containing exponential expressions, you want to group like factors together. For example:

\begin{align*} (x^3y^8)(x^{11}y^4)\amp=(x^3x^{11})(y^8y^4)\\ \amp=x^{3+11}y^{8+4}\\ \amp=x^{14}y^{12} \end{align*}

Let's make sure that the final result makes sense. In terms of \(x\text{,}\) we started with \(3\) factors and another \(11\) factors, so it makes sense that we ended up with \(14\) factors. As for \(y\text{,}\) we started with \(8\) factors and another \(4\) factors, so there are indeed \(12\) factors in all.

While you may very well be able to just count factors in your head without writing out the rule, I suggest that you write out the rule, at least at first. Down the road you will likely encounter negative exponents and even fractional exponents, and when you get there counting factors won't work. The good news is that the rules of exponents stay the same, no matter how bizarre the exponents become. So if you do forget a rule, you can recreate the rule using small positive exponents and then apply that rule to your strange exponents.

Consider the expression \((x^2)^4\text{.}\) Inside the parentheses there are two factors of \(x\text{,}\) and raising that expression to the fourth power results in \((x \cdot x)(x \cdot x)(x \cdot x)(x \cdot x)\text{.}\) So all together there are two factor of \(x\text{,}\) four times. That is, there are four times two factors of \(x\text{.}\) From this we can infer that when a power of \(x\) is raised to yet another power, the two exponents are multiplied.

The Power to a Power Rule for Exponents: \((x^m)^n=x^{m \cdot n}\)

For example:

\begin{align*} (2^3)^2\amp=2^{3 \cdot 2}\\ \amp=2^6\\ \amp=64 \end{align*}

This is verified by following the order of operations.

\begin{align*} (2^3)^2\amp=8^2\\ \amp=64 \end{align*}

While it's good practice to write out the product a few times, it's best to get in the habit of simply multiplying the exponents in your head. Writing out the rule helps with retention, but once you get into multi-step simplification problems, writing out each and every step becomes cumbersome. Look at each of the simplifications below and verify that the new exponent is the product of the two original exponents.

\((x^{12})^3=x^{36}\)

\((y^3)^7=y^{21}\)

\((w^{10})^5=w^{50}\)

Consider the expression \((xy)^3\text{.}\) The expression is short-hand for \((xy)(xy)(xy)\text{,}\) and because multiplication is both commutative and associative, the latter expression is equivalent to \((x \cdot x \cdot x)(y \cdot y \cdot y)\text{.}\) In short:

\begin{equation*} (xy)^3=x^3y^3\text{.} \end{equation*}

In words, we say that exponents distribute over multiplication.

The Product to a Power Rule of Exponents: \((xy)^n=x^ny^n\)

\begin{align*} (3 \cdot 4)^2\amp=3^2 \cdot 4^2\\ \amp=9 \cdot 16\\ \amp=144 \end{align*}
\begin{align*} (3 \cdot 4)^2\amp=12^2\\ \amp=144 \end{align*}

Similarly, exponents distribute over quotients.:

The Quotient to a Power Rule for Exponents \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\)

The new rules are supported numerically in the side-by-side examples below.

\begin{align*} \left(\frac{2}{5}\right)^3\amp=\frac{2^3}{5^3}\\ \amp=\frac{2 \cdot 2 \cdot 2}{5 \cdot 5 \cdot 5}\\ \amp=\frac{8}{125} \end{align*}
\begin{align*} \left(\frac{2}{5}\right)^3\amp=\frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5}\\ \amp=\frac{2 \cdot 2 \cdot 2}{5 \cdot 5 \cdot 5}\\ \amp=\frac{8}{125} \end{align*}

In the examples below, both the product (or quotient)to a power rule, \((xy)^n=x^ny^n\) and the power to a power rule, \((x^m)^n=x^{mn}\) are applied. Please note the the multiplication involved in the power to a power rule was done in "my head."

Example2.2.1

Simplify \((2x^4)^5\)

Solution
\begin{align*} (2x^4)^5\amp=2^5(x^4)^5\\ \amp=32x^{20} \end{align*}
Example2.2.2

Simplify \((-x^4y^2)^6\)

Solution
\begin{align*} (-x^4y^2)^6\amp=(-1)^6(x^4)^6(y^2)^6\\ \amp=1 \cdot x^{24}y^{12}\\ \amp=x^{24}y^{12} \end{align*}
Example2.2.3

Simplify \(\left(\frac{s^4}{t^7}\right)^4\)

Solution
\begin{align*} \left(\frac{s^4}{t^7}\right)^4\amp=\frac{(s^4)^4}{(t^7)^4}\\ \amp=\frac{s^{16}}{t^{28}} \end{align*}

One of the great things about the rules of exponents is that when multiple rules have to be applied, the rules can be applied in any order you like. So long as you execute each applicable rule correctly, and resist the urge to make up new rules, you should end up with the correct simplification. Compare the two paths below.

\begin{align*} \left(\frac{x^3x^6}{y^5}\right)^4\amp=\left(\frac{x^9}{y^5}\right)^4\\ \amp=\frac{(x^9)^4}{(y^5)^4}\\ \amp=\frac{x^{36}}{y^{20}} \end{align*}
\begin{align*} \left(\frac{x^3x^6}{y^5}\right)^4\amp=\frac{(x^3)^4(x^6)^4}{(y^5)^4}\\ \amp=\frac{x^{12}x^{24}}{y^{20}}\\ \amp=\frac{x^{36}}{y^{20}} \end{align*}

Let's infer another property of exponents by considering the following.

\begin{align*} \frac{x^7}{x^5}\amp=\frac{x \cdot x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x \cdot x \cdot x}\\ \amp=\frac{x}{x} \cdot \frac{x}{x} \cdot \frac{x}{x} \cdot \frac{x}{x} \cdot \frac{x}{x} \cdot \frac{x \cdot x}{1}\\ \amp=1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot x^2\\ \amp=x^2 \end{align*}

In words, five of the original seven factors of \(x\) that appeared in the numerator divided to one with the five factors of \(x\) that appeared in the denominator, leaving behind \("7-5"\) factors of \(x\) in the numerator. From this we infer the following rule.

The Quotient Rule for Exponents: \(\frac{x^m}{x^n}=x^{m-n}\)

As with the other rules, it's good to write out the step a few times to help with retention, but in the long run you'll want to perform the subtraction in your head. What follows are three examples where the step is shown, followed by two more complicated examples where the calculations (addition, multiplication, or subtraction) were done in my head.

Example2.2.4

Simplify \(\frac{w^{76}}{w^{55}}\)

Solution
\begin{align*} \frac{w^{76}}{w^{55}}\amp=w^{76-55}\\ \amp=w^{21} \end{align*}
Example2.2.5

Simplify \(\frac{a^{12}b^{21}}{a^7b^{11}}\)

Solution
\begin{align*} \frac{a^{12}b^{21}}{a^7b^{11}}\amp=a^{12-7}b^{21-11}\\ \amp=a^5b^{10} \end{align*}
Example2.2.6

Simplify \(\frac{3^8}{3^6}\)

Solution
\begin{align*} \frac{3^8}{3^6}\amp=3^{8-6}\\ \amp=3^2\\ \amp=9 \end{align*}
Example2.2.7

Simplify \(\left(\frac{x^5x^{12}}{x^4}\right)^2\)

Solution
\begin{align*} \left(\frac{x^5x^{12}}{x^4}\right)^2\amp=\left(\frac{x^{17}}{x^4}\right)^2\\ \amp=(x^{13})^2\\ \amp=x^{26} \end{align*}
Example2.2.8

Simplify \(\frac{(w^3z^5)^2}{(w^2z^2)^3}\)

Solution
\begin{align*} \frac{(w^3z^5)^2}{(w^2z^2)^3}\amp=\frac{(w^3)^2(z^5)^2}{(w^2)^3(z^2)^3}\\ \amp=\frac{w^6z^{10}}{w^6z^6}\\ \amp=z^4 \end{align*}

One last thing. Consider \(\frac{3^4}{3^4}\text{.}\) That simplifies to \(\frac{81}{81}\text{,}\) which of course is equal to \(1\text{.}\) But according to the quotient rule for exponents:

\begin{align*} \frac{3^4}{3^4}\amp=3^{4-4}\\ \amp=3^0 \end{align*}

Since \(\frac{3^4}{3^4}\) simplifies to both \(3^0\) and \(1\text{,}\) it must be the case that \(3^0=1\text{.}\) In general:

Zero Exponents: \(x^0=1, x \neq 0\)

Subsection2.2.1Exercises

Simplify each of the following expressions. The expression is not simplified unless each variable appears only once and there is only one constant factor.

1

\(x^2x^5\)

Solution

\(\begin{aligned}[t] x^2x^5\amp=x^{2+5}\\ \amp=x^7 \end{aligned}\)

2

\(a^3a\)

Solution

\(\begin{aligned}[t] a^3a\amp=a^3a^1\\ \amp=a^{3+1}\\ \amp=a^4 \end{aligned}\)

3

\(y^3y^3\)

Solution

\(\begin{aligned}[t] y^3y^3\amp=y^{3+3}\\ \amp=y^6 \end{aligned}\)

4

\(w^{27}w^{42}\)

Solution

\(\begin{aligned}[t] w^{27}w^{42}\amp=w^{27+42}\\ \amp=w^{69} \end{aligned}\)

5

\(x^{44}x^{17}\)

Solution

\(\begin{aligned}[t] x^{44}x^{17}\amp=x^{44+17}\\ \amp=x^{61} \end{aligned}\)

6

\((a^4b^2)(a^7b)\)

Solution

\(\begin{aligned}[t] (a^4b^2)(a^7b)\amp=a^{4+7}b^{2+1}\\ \amp=a^{11}b^3 \end{aligned}\)

7

\((3x^6)(-2x^5)\)

Solution

\(\begin{aligned}[t] (3x^6)(-2x^5)\amp=(3 \cdot -2)x^{6+5}\\ \amp=-6x^{11} \end{aligned}\)

8

\((4y^{12})\left(\frac{y^{12}}{4}\right)\)

Solution

\(\begin{aligned}[t] (4y^{12})\left(\frac{y^{12}}{4}\right)\amp=\left(4 \cdot \frac{1}{4}\right)y^{12+12}\\ \amp=1 \cdot y^{24}\\ \amp=y^{24} \end{aligned}\)

9

\((x^2)^3\)

Solution

\(\begin{aligned}[t] (x^2)^3\amp=x^{2 \times 3}\\ \amp=x^6 \end{aligned}\)

10

\((t^4)^2\)

Solution

\(\begin{aligned}[t] (t^4)^2\amp=t^{4 \times 2}\\ \amp=t^8 \end{aligned}\)

11

\((w^3)^2\)

Solution

\(\begin{aligned}[t] (w^3)^2\amp=w^{3 \times 2}\\ \amp=w^6 \end{aligned}\)

12

\((w^{20})^4\)

Solution

\(\begin{aligned}[t] (w^{20})^4\amp=w^{20 \times 4}\\ \amp=w^{80} \end{aligned}\)

13

\((ab)^4\)

Solution

\((ab)^4=a^4b^4\)

14

\((2a)^3\)

Solution

\(\begin{aligned}[t] (2a)^3\amp=2^3a^3\\ \amp=8a^3 \end{aligned}\)

15

\((-3t)^4\)

Solution

\(\begin{aligned}[t] (-3t)^4\amp=(-3)^4t^4\\ \amp=81t^4 \end{aligned}\)

16

\((4x^5)^3\)

Solution

\(\begin{aligned}[t] (4x^5)^3\amp=4^3(x^5)^3\\ \amp=64x^{15} \end{aligned}\)

17

\((-2w^2)^4\)

Solution

\(\begin{aligned}[t] (-2w^2)^4\amp=(-2)^4(w^2)^4\\ \amp=16w^8 \end{aligned}\)

18

\((3t^{12})(-t^{64})\)

Solution

\((3t^{12})(-t^{64})=-3t^{76}\)

19

\(\left(\frac{1}{2}r^7\right)^5\)

Solution

\(\begin{aligned}[t] \left(\frac{1}{2}r^7\right)^5\amp=\left(\frac{1}{2}\right)^5(r^7)^5\\ \amp=\frac{1}{32}r^{35} \end{aligned}\)

20

\((-2s^4t)^7\)

Solution

\(\begin{aligned}[t] (-2s^4t)^7\amp=(-2)^7(s^4)^7t^7\\ \amp=-128s^{28}t^7 \end{aligned}\)

21

\((-x^2y^4)^2\)

Solution

\(\begin{aligned}[t] (-x^2y^4)^2\amp=(-1)^2(x^2)^2(y^4)^2\\ \amp=1 \cdot x^4y^8\\ \amp=x^4y^8 \end{aligned}\)

22

\(\frac{x^5}{x^3}\)

Solution

\(\begin{aligned}[t] \frac{x^5}{x^3}\amp=x^{5-3}\\ \amp=x^2 \end{aligned}\)

23

\(\frac{2^5}{2^3}\)

Solution

\(\begin{aligned}[t] \frac{2^5}{2^3}\amp=\frac{32}{8}\\ \amp=4 \end{aligned}\)

24

\(\frac{w^{59}}{w^{36}}\)

Solution

\(\begin{aligned}[t] \frac{w^{59}}{w^{36}}\amp=w^{59-36}\\ \amp=w^{23} \end{aligned}\)

25

\(\frac{x^9}{x^9}\)

Solution

\(\begin{aligned}[t] \frac{x^9}{x^9}\amp=x^{9-9}\\ \amp=x^0\\ \amp=1 \end{aligned}\)

26

\(\left(\frac{x}{y}\right)^3\)

Solution

\(\left(\frac{x}{y}\right)^3=\frac{x^3}{y^3}\)

27

\(\left(\frac{2}{t}\right)^4\)

Solution

\(\begin{aligned}[t] \left(\frac{2}{t}\right)^4\amp=\frac{2^4}{t^4}\\ \amp=\frac{16}{t^4} \end{aligned}\)

28

\(\left(\frac{x^8}{x^5}\right)^7\)

Solution

\(\begin{aligned}[t] \left(\frac{x^8}{x^5}\right)^7\amp=(x^3)^7\\ \amp=x^{21} \end{aligned}\)

29

\(\left(-\frac{3z^5}{xy^2}\right)^4\)

Solution

\(\begin{aligned}[t] \left(-\frac{3z^5}{xy^2}\right)^4\amp=(-1)^4 \cdot \frac{3^4(z^5)^4}{x^4(y^2)^4}\\ \amp=1 \cdot \frac{81z^{20}}{x^4y^8}\\ \amp=\frac{81z^{20}}{x^4y^8} \end{aligned}\)

30

\(\left(\frac{2^{11}x^4}{y^{77}z^{12}}\right)^0\)

Solution

\(\left(\frac{2^{11}x^4}{y^{77}z^{12}}\right)^0=1\)