Exponent Laws

An exponent is shorthand for repeated multiplication: $2^5$ means five $2$s multiplied together. The exponent laws are shortcuts for multiplying and dividing powers without writing everything out, and they also tell you what an exponent of zero or a negative exponent has to mean. You’ll use them all the time when you expand and factor polynomials later in this unit.

Key ideas

Powers

In the power $a^n$, the number $a$ is the base and $n$ is the exponent. For a positive whole number $n$,

$$a^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ factors}}$$

Watch the brackets with negative bases: $(-3)^2 = (-3)(-3) = 9$, but $-3^2 = -(3 \times 3) = -9$. The exponent only applies to what it’s touching.

The three main laws

Each law comes from counting factors. For example, $2^3 \times 2^4 = (2 \times 2 \times 2)(2 \times 2 \times 2 \times 2) = 2^7$: three factors plus four factors make seven.

Law	Rule	Example
product	$a^m \times a^n = a^{m + n}$	$5^2 \times 5^6 = 5^8$
quotient	$a^m \div a^n = a^{m - n}$, $a \ne 0$	$7^9 \div 7^4 = 7^5$
power of a power	$(a^m)^n = a^{mn}$	$(3^2)^5 = 3^{10}$

The bases must be the same for the product and quotient laws. You can’t combine $2^3 \times 5^4$ into a single power.

Power of a product or a quotient

An exponent outside brackets applies to every factor inside:

$$(ab)^n = a^n b^n \qquad\qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}, \ b \ne 0$$

For example, $(2x)^3 = 2^3 x^3 = 8x^3$ and $\left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}$.

This only works for products and quotients, not sums: $(a + b)^2$ is not $a^2 + b^2$. (See polynomial operations for how to expand $(a + b)^2$ correctly.)

Zero exponents: a pattern in y = 2^x

Look at a table of values for $y = 2^x$. Each time $x$ goes down by $1$, $y$ is divided by $2$:

$x$	$3$	$2$	$1$	$0$	$-1$	$-2$	$-3$
$y = 2^x$	$8$	$4$	$2$	$1$	$\dfrac{1}{2}$	$\dfrac{1}{4}$	$\dfrac{1}{8}$

Keeping the pattern going, $2^0 = 2 \div 2 = 1$. The quotient law agrees: $\dfrac{5^3}{5^3} = 5^{3 - 3} = 5^0$, but anything divided by itself is $1$. So for any base $a \ne 0$:

$$a^0 = 1$$

Negative exponents

The same pattern gives $2^{-1} = \dfrac{1}{2}$, $2^{-2} = \dfrac{1}{4}$ and $2^{-3} = \dfrac{1}{8}$. The quotient law agrees again:

$$\frac{2^3}{2^5} = 2^{3 - 5} = 2^{-2} \qquad \text{and} \qquad \frac{2^3}{2^5} = \frac{2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{2^2}$$

So a negative exponent means “one over the positive power”, for any base $a \ne 0$:

$$a^{-n} = \frac{1}{a^n} \qquad\qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$$

A negative exponent makes a reciprocal, not a negative number: $2^{-3} = \dfrac{1}{8}$, which is positive.

All the laws in the table above still work with zero and negative exponents. When you simplify, write your final answer with positive exponents unless you’re asked otherwise.

In this course the exponents are always integers. In Grade 11 you’ll see that the same laws work for fractional exponents too, like $8^{\frac{2}{3}}$, in rational exponents.

Comparing y = x² and y = 2^x

Both relations grow, but in very different ways. In $y = x^2$ the variable is the base; in $y = 2^x$ the variable is the exponent.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$x^2$	$4$	$1$	$0$	$1$	$4$	$9$	$16$	$25$	$36$
$2^x$	$\dfrac{1}{4}$	$\dfrac{1}{2}$	$1$	$2$	$4$	$8$	$16$	$32$	$64$

$y = x^2$ (blue) and $y = 2^x$ (orange). Each $y$-axis grid line is $2$ units.

• $y = x^2$ is a parabola: symmetric about the $y$-axis, with its lowest point at $(0, 0)$.
• $y = 2^x$ is not symmetric. Its $y$-intercept is $1$ (because $2^0 = 1$), it’s always positive, and to the left it gets closer and closer to the $x$-axis without ever touching it.
• The curves cross three times: at $(2, 4)$, at $(4, 16)$, and once for a negative $x$ (about $x = -0.77$).
• For large $x$, $2^x$ wins easily, because each step doubles it. At $x = 10$, $x^2 = 100$ but $2^{10} = 1024$.

You’ll study $y = 2^x$ and other exponential relations properly in Grade 11.

Worked examples

Example 1: Using the laws

Write each as a single power, then evaluate (d).

• (a) $3^4 \times 3^5$
• (b) $7^8 \div 7^3$
• (c) $(2^3)^4$
• (d) $(2 \times 5)^3$

Solution.

(a) Same base, multiplying: add the exponents. $3^4 \times 3^5 = 3^{4 + 5} = 3^9$

(b) Same base, dividing: subtract the exponents. $7^8 \div 7^3 = 7^{8 - 3} = 7^5$

(c) Power of a power: multiply the exponents. $(2^3)^4 = 2^{3 \times 4} = 2^{12}$

(d) Power of a product: $(2 \times 5)^3 = 2^3 \times 5^3 = 8 \times 125 = 1000$.

Check: $(2 \times 5)^3 = 10^3 = 1000$. ✓

Example 2: Zero and negative exponents

Evaluate.

• (a) $5^0$
• (b) $4^{-2}$
• (c) $(-3)^{-3}$
• (d) $\left(\dfrac{2}{3}\right)^{-2}$
• (e) $3^{-1} + 3^0$

Solution.

(a) Any non-zero base to the exponent $0$ is $1$: $5^0 = 1$.

(b) $4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16}$

(c) $(-3)^{-3} = \dfrac{1}{(-3)^3} = \dfrac{1}{-27} = -\dfrac{1}{27}$. The answer is negative because the base is negative, not because the exponent is.

(d) Flip the fraction and make the exponent positive:

$$\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$

(e) $3^{-1} + 3^0 = \dfrac{1}{3} + 1 = \dfrac{4}{3}$

Example 3: Simplifying, then evaluating

Simplify, then evaluate or write with positive exponents.

• (a) $\dfrac{2^5 \times 2^{-3}}{2^4}$
• (b) $(3^{-2})^{-3} \div 3^4$
• (c) $\dfrac{(2x^3y)^2 \times 3x^{-4}}{6xy^3}$

Solution. Simplify first using the laws. It’s much easier than evaluating each power.

(a)

$$\frac{2^5 \times 2^{-3}}{2^4} = \frac{2^{5 + (-3)}}{2^4} = \frac{2^2}{2^4} = 2^{2 - 4} = 2^{-2} = \frac{1}{4}$$

(b)

$$(3^{-2})^{-3} \div 3^4 = 3^{(-2)(-3)} \div 3^4 = 3^6 \div 3^4 = 3^2 = 9$$

(c) Square every factor in the bracket first, then deal with numbers and each variable separately:

$$\begin{aligned} \frac{(2x^3y)^2 \times 3x^{-4}}{6xy^3} &= \frac{4x^6y^2 \times 3x^{-4}}{6xy^3} && \text{power of a product} \\ &= \frac{12x^{2}y^2}{6xy^3} && \text{product law: } 6 + (-4) = 2 \\ &= 2x^{2 - 1}y^{2 - 3} && \text{quotient law} \\ &= 2xy^{-1} \\ &= \frac{2x}{y} \end{aligned}$$

Check (c) with $x = 1$, $y = 2$: the original is $\dfrac{(4)^2 \times 3}{6 \times 8} = \dfrac{48}{48} = 1$, and $\dfrac{2x}{y} = \dfrac{2}{2} = 1$. ✓

Example 4: Comparing x² and 2^x

Use the table in Key ideas.

• (a) For which values of $x$ in the table are $x^2$ and $2^x$ equal?
• (b) At $x = 3$, which is bigger? At $x = 6$?
• (c) Which relation has a $y$-value of $1$ when $x = 0$? Explain using an exponent law.

Solution.

(a) They’re equal at $x = 2$ (both $4$) and at $x = 4$ (both $16$). These are the crossing points $(2, 4)$ and $(4, 16)$ on the graph.

(b) At $x = 3$: $x^2 = 9$ and $2^x = 8$, so $x^2$ is bigger. At $x = 6$: $x^2 = 36$ and $2^x = 64$, so $2^x$ is bigger, and it stays ahead from $x = 5$ onward.

(c) $y = 2^x$, because $2^0 = 1$ (any non-zero base to the exponent $0$ is $1$). For $y = x^2$, $0^2 = 0$.

Common mistakes

Multiplying the bases. $2^3 \times 2^4$ is $2^7$, not $4^7$. The base stays the same; only the exponents combine.

Multiplying exponents instead of adding them. $x^2 \times x^3 = x^5$, not $x^6$. Multiply exponents only for a power of a power, like $(x^2)^3 = x^6$. If in doubt, write out the factors.

Thinking a negative exponent makes a negative number. $2^{-3} = \dfrac{1}{8}$, not $-8$. A negative exponent means “reciprocal”.

Saying a zero exponent gives zero. $5^0 = 1$, not $0$. Look back at the $2^x$ table: the pattern $8, 4, 2, \ldots$ continues to $1$, not $0$.

Forgetting to apply the exponent to the number. $(2x)^3 = 8x^3$, not $2x^3$. The exponent applies to every factor inside the brackets. On the other hand, in $2x^3$ (no brackets) only the $x$ is cubed.

Mixing up the negative base and the negative sign. $(-4)^2 = 16$, but $-4^2 = -16$. Likewise $-4^0 = -1$, because only the $4$ is raised to the exponent $0$.

Practice

1. (Warm-up) Write each as a single power.

• (a) $6^3 \times 6^7$
• (b) $9^{10} \div 9^4$
• (c) $(5^2)^6$

Solution

(a) $6^{3 + 7} = 6^{10}$

(b) $9^{10 - 4} = 9^6$

(c) $5^{2 \times 6} = 5^{12}$

2. (Warm-up) Evaluate.

• (a) $8^0$
• (b) $2^{-5}$
• (c) $10^{-3}$
• (d) $-4^0$

Solution

(a) $8^0 = 1$

(b) $2^{-5} = \dfrac{1}{2^5} = \dfrac{1}{32}$

(c) $10^{-3} = \dfrac{1}{10^3} = \dfrac{1}{1000} = 0.001$

(d) Only the $4$ has the exponent: $-4^0 = -(4^0) = -1$.

3. (Warm-up) Complete a table of values for $y = 3^x$ for $x = 2, 1, 0, -1, -2, -3$. Describe the pattern, and use it to explain why $3^0 = 1$.

Solution

$x$	$2$	$1$	$0$	$-1$	$-2$	$-3$
$y = 3^x$	$9$	$3$	$1$	$\dfrac{1}{3}$	$\dfrac{1}{9}$	$\dfrac{1}{27}$

Each time $x$ goes down by $1$, $y$ is divided by $3$. Going from $x = 1$ to $x = 0$ gives $3 \div 3 = 1$, so $3^0 = 1$.

4. (Core) Evaluate.

• (a) $(-2)^{-3}$
• (b) $\left(\dfrac{3}{4}\right)^{-2}$
• (c) $4^{-1} + 2^{-2}$

Solution

(a) $(-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = -\dfrac{1}{8}$

(b) $\left(\dfrac{3}{4}\right)^{-2} = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}$

(c) $4^{-1} + 2^{-2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$

5. (Core) Simplify, then evaluate.

• (a) $\dfrac{5^7 \times 5^{-4}}{5^5}$
• (b) $(2^{-3})^2 \times 2^8$

Solution

(a)

$$\frac{5^7 \times 5^{-4}}{5^5} = \frac{5^3}{5^5} = 5^{-2} = \frac{1}{25}$$

(b)

$$(2^{-3})^2 \times 2^8 = 2^{-6} \times 2^8 = 2^2 = 4$$

6. (Core) Simplify. Write your answers with positive exponents.

• (a) $(3x^2)^3$
• (b) $\dfrac{12a^5b^2}{4a^7b}$
• (c) $(x^{-2}y^3)^{-2}$

Solution

(a) $(3x^2)^3 = 3^3 x^{2 \times 3} = 27x^6$

(b)

$$\frac{12a^5b^2}{4a^7b} = 3a^{5 - 7}b^{2 - 1} = 3a^{-2}b = \frac{3b}{a^2}$$

(c)

$$(x^{-2}y^3)^{-2} = x^{(-2)(-2)}y^{3(-2)} = x^4y^{-6} = \frac{x^4}{y^6}$$

7. (Core) Use the quotient law to explain why $7^0 = 1$ and why $7^{-2} = \dfrac{1}{49}$.

Solution

By the quotient law, $\dfrac{7^4}{7^4} = 7^{4 - 4} = 7^0$. But any non-zero number divided by itself is $1$, so $7^0 = 1$.

By the quotient law, $\dfrac{7^3}{7^5} = 7^{3 - 5} = 7^{-2}$. Writing out the factors, three $7$s cancel from the top and bottom:

$$\frac{7^3}{7^5} = \frac{7 \times 7 \times 7}{7 \times 7 \times 7 \times 7 \times 7} = \frac{1}{7 \times 7} = \frac{1}{49}$$

So $7^{-2} = \dfrac{1}{49}$.

8. (Challenge) A bacteria culture in a lab doubles every hour. Right now it has $500$ bacteria, so the number $N$ after $t$ hours is $N = 500 \times 2^t$.

• (a) How many bacteria will there be in $3$ hours?
• (b) What does $t = 0$ give? Does that make sense?
• (c) What does $t = -2$ mean? Find $N$ for $t = -2$.

Solution

(a) $N = 500 \times 2^3 = 500 \times 8 = 4000$ bacteria.

(b) $N = 500 \times 2^0 = 500 \times 1 = 500$. That’s the number right now, which makes sense: $t = 0$ is “now”.

(c) $t = -2$ means $2$ hours ago.

$$N = 500 \times 2^{-2} = 500 \times \frac{1}{4} = 125$$

There were $125$ bacteria $2$ hours ago. Check: doubling twice, $125 \to 250 \to 500$. ✓

9. (Challenge) For which positive whole numbers $x$ is $x^2$ greater than $2^x$? Use a table, then explain why $2^x$ stays ahead once $x$ is $5$ or more.

Solution

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$x^2$	$1$	$4$	$9$	$16$	$25$	$36$	$49$
$2^x$	$2$	$4$	$8$	$16$	$32$	$64$	$128$

Only at $x = 3$ is $x^2$ greater ($9 \gt 8$). At $x = 2$ and $x = 4$ they’re equal.

Why $2^x$ stays ahead: each time $x$ goes up by $1$, $2^x$ doubles. But $x^2$ grows by much less than double once $x$ is $5$ or more. For example, from $x = 5$ to $x = 6$, $x^2$ goes from $25$ to $36$, which is multiplied by only $1.44$. The bigger $x$ gets, the closer that multiplier is to $1$. Since $2^x$ is already ahead at $x = 5$ and grows faster at every step after that, $x^2$ can never catch up.