Laws of Exponents

Learning goals

Add exponents to multiply powers of the same base
Subtract exponents to divide powers of the same base
Multiply exponents when raising a power to a power
Spread an exponent across a product or a quotient
Derive each law by counting copies of the base
Refuse to apply a law to $a^m + a^n$ , which has none

The one definition everything rests on

Every rule below is read straight off the definition of a power from the previous lesson. Start with a case you can check by hand: $3^4$ is $3 \times 3 \times 3 \times 3$ , which multiplies out to $81$ . Counting copies of the base works the same way for any base $a$ and whole-number exponent $n \ge 1$ ,

a^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ copies of } a}.

Each law below counts these copies in the same way, when powers are multiplied, divided, or raised to another power. Throughout this lesson the exponents are whole numbers ( $1$ or more), and the bases are ordinary numbers you have already met.

The product rule: multiply by adding exponents

Suppose you multiply two powers that share the same base, say $2^3 \times 2^4$ . Replace each power by its factors and count what you get:

2^3 \times 2^4 = \underbrace{(2 \times 2 \times 2)}_{3 \text{ factors}} \times \underbrace{(2 \times 2 \times 2 \times 2)}_{4 \text{ factors}} = \underbrace{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}_{7 \text{ factors}} = 2^7.

Three $2$ ‘s sitting next to four $2$ ‘s is just seven $2$ ‘s in a row, so the answer is $2^7$ . Notice that $3 + 4 = 7$ : the exponent of the answer is the sum of the two exponents.

The product rule: $a^m \times a^n = a^{m+n}$ #

Take $6^2 \times 6^3$ , two powers that share the base $6$ . The first power is two copies of $6$ and the second is three copies of $6$ . Setting them side by side gives one run of five $6$ ‘s:

6^2 \times 6^3 = \underbrace{(6 \times 6)}_{2} \times \underbrace{(6 \times 6 \times 6)}_{3} = \underbrace{6 \times 6 \times 6 \times 6 \times 6}_{5} = 6^5.

The run holds $2 + 3 = 5$ factors of $6$ , which is exactly why the two exponents add.

Nothing in that count depended on the particular numbers $6$ , $2$ and $3$ , so run it again with letters. Take any base $a$ and two whole-number exponents $m$ and $n$ . By the definition of a power, $a^m$ is $m$ factors of $a$ multiplied together, and $a^n$ is $n$ factors of $a$ multiplied together.

When you multiply $a^m$ by $a^n$ , you set those two groups of factors side by side in one long product. Multiplication lets you drop the inner grouping, so nothing separates them:

a^m \times a^n = \underbrace{(a \times \cdots \times a)}_{m} \times \underbrace{(a \times \cdots \times a)}_{n} = \underbrace{a \times \cdots \times a}_{m + n}.

The combined product is one run of $a$ ‘s, and its length is $m$ copies followed by $n$ copies, which is $m + n$ copies in all. A product of $m + n$ factors of $a$ is, by the same definition, $a^{m+n}$ .

In words: to multiply powers of the same base, keep the base and add the exponents.

Worked example 1 Write $5^2 \times 5^6$ as a single power

Both powers share the base $5$ , so the product rule applies: keep the base and add the exponents.

5^2 \times 5^6 = 5^{2+6} = 5^8.

If you ever doubt it, expand: two $5$ ‘s times six $5$ ‘s is eight $5$ ‘s in a row, which is $5^8$ . You would only multiply it out to a plain number if the problem asked you to evaluate it.

The same rule handles a string of three or more powers, because you can combine them two at a time. So $3^2 \times 3^3 \times 3^4 = 3^{2+3+4} = 3^9$ : add up every exponent at once. Since the exponents add, you can split a power apart whenever it helps, writing $2^7$ as $2^3 \times 2^4$ or as $2^5 \times 2^2$ .

Check your understanding

Write $4^5 \times 4^3$ as a single power of $4$ .

4^8

4^{15}

16^8

4^2

The quotient rule: divide by subtracting exponents

Division undoes multiplication, so dividing two powers of the same base takes copies of the base away instead of piling them up. Look at $\dfrac{2^5}{2^3}$ and expand the top and bottom into factors:

\frac{2^5}{2^3} = \frac{2 \times 2 \times 2 \times 2 \times 2}{2 \times 2 \times 2}.

Every $2$ on the bottom cancels a $2$ on the top, because $\tfrac{2}{2} = 1$ . Three factors cancel from each, leaving

\frac{2^5}{2^3} = \frac{\cancel{2} \times \cancel{2} \times \cancel{2} \times 2 \times 2}{\cancel{2} \times \cancel{2} \times \cancel{2}} = 2 \times 2 = 2^2.

Five factors with three cancelled leaves $5 - 3 = 2$ of them, so the answer is $2^2$ . The exponent of the answer is the difference of the two exponents.

The quotient rule: $\dfrac{a^m}{a^n} = a^{m-n}$ when $m > n$ and $a \neq 0$ #

Start with $\dfrac{4^6}{4^2}$ , where the base $4$ is shared and is not zero. The top is six copies of $4$ and the bottom is two copies of $4$ , and each factor below cancels one above since $\tfrac{4}{4} = 1$ :

\frac{4^6}{4^2} = \frac{\cancel{4} \times \cancel{4} \times 4 \times 4 \times 4 \times 4}{\cancel{4} \times \cancel{4}} = 4 \times 4 \times 4 \times 4 = 4^4.

Two of the six factors on top are used up, so $6 - 2 = 4$ of them survive, which is why the exponents subtract.

Nothing in those steps depended on the particular numbers $4$ , $6$ and $2$ , so run the same argument with letters. Take a base $a$ that is not zero (so the division makes sense) and two whole-number exponents with $m > n$ . Write the quotient with each power expanded into factors. The numerator is $m$ copies of $a$ and the denominator is $n$ copies of $a$ :

\frac{a^m}{a^n} = \frac{\overbrace{a \times \cdots \times a}^{m}}{\underbrace{a \times \cdots \times a}_{n}}.

Each factor of $a$ in the denominator cancels one factor of $a$ in the numerator, since $\tfrac{a}{a} = 1$ . There are $n$ factors below, so $n$ of the factors above are cancelled. Because $m$ is larger than $n$ , there are factors left over on top. The number left over is $m - n$ :

\frac{a^m}{a^n} = \underbrace{a \times \cdots \times a}_{m - n} = a^{m-n}.

A product of $m - n$ factors of $a$ is $a^{m-n}$ by the definition, which proves the rule.

In words: to divide powers of the same base, keep the base and subtract the exponent of the denominator from the exponent of the numerator. Notice the proof needed $m > n$ , so that something is left after cancelling, and it needed $a \neq 0$ , so that $\tfrac{a}{a} = 1$ is honest. For now we keep $m > n$ , which always leaves a positive exponent.

Worked example 2 Simplify $\dfrac{7^9}{7^4}$

The base $7$ is shared, and the top exponent is larger than the bottom one, so subtract.

\frac{7^9}{7^4} = 7^{9-4} = 7^5.

To see it directly, nine $7$ ‘s over four $7$ ‘s cancels four pairs, leaving five $7$ ‘s on top, which is $7^5$ .

The quotient rule subtracts exponents, but what if the bottom exponent equals the top one, as in $\tfrac{2^3}{2^3}$ , or is even larger? Subtracting would give an exponent of $0$ or a negative number. So far a power has only meant “multiply the base this many times,” which makes no sense for $0$ or a negative count. That is a real and interesting gap, and the very next lesson, on zero and negative exponents, is devoted to filling it.

Check your understanding

Simplify $\dfrac{6^8}{6^2}$ to a single power of $6$ .

6^6

6^4

6^{10}

1^6

The power rule: a power of a power multiplies exponents

The third situation is a power raised to another power, like $(2^3)^2$ . The outer exponent $2$ says to use $2^3$ as a factor twice, so write it out and then count the factors of $2$ :

(2^3)^2 = 2^3 \times 2^3 = \underbrace{(2 \times 2 \times 2)}_{3} \times \underbrace{(2 \times 2 \times 2)}_{3} = \underbrace{2 \times 2 \times 2 \times 2 \times 2 \times 2}_{6} = 2^6.

Two groups of three factors each is $2 \times 3 = 6$ factors in all, so the answer is $2^6$ . This time the exponents multiply.

The power rule: $(a^m)^n = a^{m \times n}$ #

Start with $(7^2)^5$ , a power that is itself raised to a power. The outer exponent $5$ says to use the quantity $7^2$ as a factor five times. Each of those five copies is itself two factors of $7$ :

\begin{aligned} (7^2)^5 &= \underbrace{(7 \times 7)}_{2} \times \underbrace{(7 \times 7)}_{2} \times \underbrace{(7 \times 7)}_{2} \\ &\qquad \times \underbrace{(7 \times 7)}_{2} \times \underbrace{(7 \times 7)}_{2} \\ &= \underbrace{7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7}_{10} \\ &= 7^{10}. \end{aligned}

Five groups of two factors each hold $2 \times 5 = 10$ factors of $7$ altogether, which is why these exponents multiply rather than add.

Nothing in that count depended on the particular numbers $7$ , $2$ and $5$ , so run it again with letters. Take a base $a$ and whole-number exponents $m$ and $n$ . The outer exponent $n$ means to use the quantity $a^m$ as a factor $n$ times:

(a^m)^n = \underbrace{a^m \times a^m \times \cdots \times a^m}_{n \text{ copies of } a^m}.

Each one of those copies is itself $m$ factors of $a$ . So you have $n$ groups, and every group holds $m$ factors of $a$ . The total number of factors of $a$ is therefore $m$ added to itself $n$ times, which is $m \times n$ :

(a^m)^n = \underbrace{(\,\underbrace{a \cdots a}_{m}\,)(\,\underbrace{a \cdots a}_{m}\,)\cdots(\,\underbrace{a \cdots a}_{m}\,)}_{n \text{ groups}} = \underbrace{a \times \cdots \times a}_{m \times n}.

A product of $m \times n$ factors of $a$ is $a^{m \times n}$ , so $(a^m)^n = a^{m \times n}$ .

In words: to raise a power to a power, keep the base and multiply the two exponents. This is the rule most often confused with the product rule, so hold the two apart: in $2^3 \times 2^4$ the powers sit side by side. There the exponents add to give $2^7$ , while in $(2^3)^4$ one power is raised to another and the exponents multiply to give $2^{12}$ .

Worked example 3 Simplify $(10^4)^3$

This is a power raised to a power, so multiply the exponents.

(10^4)^3 = 10^{4 \times 3} = 10^{12}.

Reading it the long way confirms the count: $10^4$ used three times is three groups of four factors of $10$ . Those groups hold twelve factors in all, so the answer is $10^{12}$ .

Spreading a power across a product or a quotient

The last two laws are about what happens when the base is itself a product or a quotient. Consider $(2 \times 5)^3$ . The exponent $3$ asks for three copies of the base $2 \times 5$ multiplied together. Because multiplication can be reordered freely (a property you met early on), you can gather all the $2$ ‘s and all the $5$ ‘s:

(2 \times 5)^3 = (2 \times 5)(2 \times 5)(2 \times 5) = \underbrace{(2 \times 2 \times 2)}_{2^3} \times \underbrace{(5 \times 5 \times 5)}_{5^3} = 2^3 \times 5^3.

So the exponent lands on each factor separately. The same regrouping works for a quotient, since a fraction raised to a power multiplies that many copies of the fraction. Check that on $\left(\dfrac{2}{3}\right)^2$ , which is $\dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{4}{9}$ , and $\dfrac{4}{9}$ is exactly $\dfrac{2^2}{3^2}$ .

Power of a product and of a quotient: $(ab)^n = a^n b^n$ and $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$ #

Take $(3 \times 7)^2$ , a power whose base is itself a product. The exponent $2$ asks for two copies of $3 \times 7$ multiplied together. Multiplication can be reordered, so the two $3$ ‘s gather together and the two $7$ ‘s gather together:

(3 \times 7)^2 = (3 \times 7)(3 \times 7) = \underbrace{(3 \times 3)}_{3^2} \times \underbrace{(7 \times 7)}_{7^2} = 9 \times 49 = 441.

The base $3 \times 7$ is $21$ , and $21^2$ is also $441$ , so the exponent landed on each factor separately.

Nothing in that regrouping depended on the particular numbers $3$ , $7$ and $2$ , so run it again with letters. By the definition, $(ab)^n$ is $n$ copies of $ab$ multiplied together:

(ab)^n = \underbrace{(ab)(ab)\cdots(ab)}_{n \text{ copies}}.

Multiplication can be reordered and regrouped however you like. So collect the $n$ copies of $a$ into one group and the $n$ copies of $b$ into another:

(ab)^n = \underbrace{(a \cdots a)}_{n} \times \underbrace{(b \cdots b)}_{n} = a^n b^n.

The quotient works the same way, with $b \neq 0$ . On numbers, $\left(\tfrac{4}{5}\right)^3$ multiplies three copies of $\tfrac{4}{5}$ , which puts three $4$ ‘s on top and three $5$ ‘s on the bottom and gives $\tfrac{64}{125}$ . Raising $\tfrac{a}{b}$ to the $n$ multiplies $n$ copies of the fraction, and multiplying fractions multiplies the numerators and the denominators separately. So $n$ factors of $a$ collect on top and $n$ factors of $b$ collect on the bottom:

\left(\frac{a}{b}\right)^n = \underbrace{\frac{a}{b} \times \cdots \times \frac{a}{b}}_{n} = \frac{\overbrace{a \times \cdots \times a}^{n}}{\underbrace{b \times \cdots \times b}_{n}} = \frac{a^n}{b^n}.

In words: a power of a product is the product of the powers. Likewise a power of a quotient is the quotient of the powers, provided the denominator is not zero. This is what lets you rewrite something like $6^3$ by splitting the base: since $6 = 2 \times 3$ , you get $6^3 = (2 \times 3)^3 = 2^3 \times 3^3 = 8 \times 27$ .

Each law just rebundles the same factors: powers side by side add their counts, a power of a power multiplies them, and a power of a product splits onto each factor.

Check your understanding

Which expression equals $(3 \times 4)^2$ ?

3^2 \times 4^2

3^2 + 4^2

3 \times 4^2

12 \times 2

Putting the laws together

Most real simplifications use more than one law in turn. Work from the inside out, just as the order of operations tells you, resolving any power-of-a-power before you combine side-by-side powers.

Worked example 4 Simplify $\dfrac{(3^2)^4}{3^5}$ to a single power of $3$

Resolve the power of a power on top first, using the power rule (multiply the exponents):

(3^2)^4 = 3^{2 \times 4} = 3^8.

The expression is now $\dfrac{3^8}{3^5}$ , a single base over a single base, so apply the quotient rule (subtract the exponents):

\frac{3^8}{3^5} = 3^{8-5} = 3^3.

So $\dfrac{(3^2)^4}{3^5} = 3^3$ .

Worked example 5 Simplify $2^3 \times 2^4 \times 5^2$ as far as the laws allow

The two powers of $2$ share a base, so the product rule combines them:

2^3 \times 2^4 = 2^{3+4} = 2^7.

That leaves $2^7 \times 5^2$ . The bases $2$ and $5$ are different, so the product rule cannot pool them. The law that does reach across different bases is the power of a product read backwards, $a^n \times b^n = (ab)^n$ , and it needs the same exponent on each. Here the exponents $7$ and $2$ differ, so the two complete powers cannot be combined into a single power of one base. You can still regroup the factors, since $2^7 \times 5^2 = 2^5 \times (2^2 \times 5^2) = 2^5 \times 10^2$ . That trades one prime base for a composite one, so the prime-power form $2^7 \times 5^2$ is the one to keep:

2^3 \times 2^4 \times 5^2 = 2^7 \times 5^2.

You could go further only by evaluating to a plain number, $128 \times 25 = 3200$ .

Common mistakes

Multiplying the exponents when you should add them. For a product of powers, the exponents add: $2^3 \times 2^4 = 2^7$ , not $2^{12}$ . Multiplying exponents is the rule for a power of a power, $(2^3)^4 = 2^{12}$ , a different setup entirely.
Multiplying the bases too. $2^3 \times 2^4 = 2^7$ , not $4^7$ . The base stays the same when you combine powers; only the count of factors changes. Check it on a tiny case: $2^1 \times 2^1 = 4$ is $2^2$ , not $4^2 = 16$ .
Trying to combine different bases with the product rule. The product and quotient rules require the same base, so $2^3 \times 5^2$ stays $2^3 \times 5^2$ . A shared exponent is a different matter: $2^3 \times 5^3 = (2 \times 5)^3 = 10^3$ .
Using an exponent law on a sum. The laws are about multiplying and dividing, never adding. So $2^3 + 2^4$ does not become $2^7$ or any other single power; it is just $8 + 16 = 24$ . Likewise $a^m + a^n$ has no exponent-law shortcut.
Splitting a power across a sum. A power distributes over a product, but not over a sum: $(2 + 3)^2 = 5^2 = 25$ , which is not $2^2 + 3^2 = 13$ . Only $(2 \times 3)^2 = 2^2 \times 3^2$ is valid.

Key takeaways

Product rule: same base, multiply by adding exponents: $a^m \times a^n = a^{m+n}$ .
Quotient rule: same base, which must not be zero, divide by subtracting exponents: $\dfrac{a^m}{a^n} = a^{m-n}$ (kept here to $m > n$ , so the exponent stays positive).
Power rule: a power of a power multiplies the exponents: $(a^m)^n = a^{m \times n}$ .
Power of a product or quotient: the exponent lands on each factor: $(ab)^n = a^n b^n$ and $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$ when $b$ is not zero.
Every one of these is just counting factors of the base; expand both sides and the rule is plain.
The product and quotient rules need a shared base, and they are for multiplying and dividing only: there is no exponent law for $a^m + a^n$ .

Laws of Exponents

Learning goals

The one definition everything rests on

The product rule: multiply by adding exponents

The quotient rule: divide by subtracting exponents

The power rule: a power of a power multiplies exponents

Spreading a power across a product or a quotient

Putting the laws together

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)