Index laws — the rules for powers — Mathematics

Before this

Index notation and the basic laws · coming soon
Standard form
Calculating with standard form

This leads to

Negative and fractional indices
What is a logarithm
The power rule in calculus · coming soon

Five, multiplied by itself no times

Here is a question that breaks the way most people first learn powers.

What is $5^{0}$ ?

Many students answer 0. Their reason sounds sensible. $5^{3}$ means three fives multiplied together. So $5^{0}$ means no fives multiplied together, and "nothing" sounds like 0.

Now type $5^{0}$ into a calculator. It says 1.

So either the calculator is wrong, or the meaning "multiply the number by itself" is not the whole story. It is not the whole story. "Multiply 5 by itself zero times" has no meaning at all. So $5^{0}$ needs a meaning that nobody has given it yet.

You have probably been told the rules for powers, and told to remember that anything to the power zero is 1. We are not going to do that. This lecture builds every rule from one definition, by counting. Then it shows that the 1 on the calculator is not a rule someone chose. It is the only answer that keeps the rules working.

One definition, and everything comes from it

For a positive whole number $n$ , the power $a^{n}$ means $n$ copies of $a$ multiplied together:

a^{n} = n factors a \times a \times \dots \times a

The number $a$ is the base. The small raised number $n$ is the index. The plural of index is indices. Some books say "exponent" instead of index. You will also hear "power": $a^{n}$ is read " $a$ to the power $n$ ".

Each copy of $a$ in the product is called a factor. So the index simply counts factors. $2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$ has five factors of 2. It is not $2 \times 5$ . That is a different question with a different answer, 10.

This definition works for $n = 1, 2, 3, \dots$ , because you can count the factors. It says nothing about $n = 0$ , because there is nothing to count. Hold on to that. It is the key to the end of the lecture.

René Descartes used raised indices such as $x^{3}$ and $x^{4}$ throughout his book La Géométrie in 1637, although he still wrote $xx$ for a square. Before that, powers were usually written with words or abbreviations.

Multiplying powers: count the factors

Take $a^{3} \times a^{4}$ . Write each power out as its factors:

a^{3} \times a^{4} = (a \times a \times a) \times (a \times a \times a \times a)

Multiplication does not care about brackets. So this is one long product of $a$ 's. Count them: 3 and then 4 more makes 7. So $a^{3} \times a^{4} = a^{7}$ .

Top: three factors of

a

times four more factors make seven factors, so

a^{3} \times a^{4} = a^{7}

. Bottom: in

a^{5} \div a^{2}

, each

a

on the bottom cancels one

a

on top, because

a \div a = 1

. Three factors are left, so

a^{5} \div a^{2} = a^{3}

Nothing in this argument depended on 3 and 4. With $m$ factors and then $n$ more, there are $m + n$ factors altogether. This gives the first law.

a^{m} \times a^{n} = a^{m + n}

The lecture on calculating with standard form used this rule for powers of ten: $1 0^{8} \times 1 0^{7} = 1 0^{15}$ , because eight tens and seven more tens make fifteen tens. It promised that a later lecture would prove the rule for any base. Here is that proof. The counting works the same way whether the base is 10, 2, $x$ or any other number.

One condition hides inside the argument. Both powers must have the same base. In $2^{3} \times 3^{4}$ , the factors are three 2s and four 3s. They are not copies of one number, so they cannot be counted together.

Dividing powers: cancel the factors

Now take $a^{5} \div a^{2}$ . Write it as a fraction, with the factors on top and underneath:

\frac{a ^{5}}{a ^{2}} = \frac{a \times a \times a \times a \times a}{a \times a}

Each $a$ underneath can be paired with one $a$ on top. Each pair is $a \div a$ , which is 1. So the pair cancels. Two pairs cancel, and three factors of $a$ are left on top. So $a^{5} \div a^{2} = a^{3}$ , which is $a^{5 - 2}$ . The bottom picture in the figure shows the cancelling.

In general, $n$ factors underneath cancel $n$ of the $m$ factors on top. This leaves $m - n$ . This is the second law.

a^{m} \div a^{n} = a^{m - n}

Read the argument again, slowly, because it has two conditions. First, the base $a$ must not be 0, because the cancelling uses $a \div a = 1$ , and $0 \div 0$ has no meaning. Second, there must be more factors on top than underneath, so $m$ must be bigger than $n$ . If $m$ and $n$ are equal, or if $m$ is smaller, the counting runs out of factors. Then the definition alone cannot say what the answer means. The explorer below lets you see this happen.

Explore — the counting machine

Choose multiply, divide or a power of a power. Then set the two indices. Each tile is one factor of a. Count the tiles, and compare your count with the law. Then try dividing with the two indices equal, and with the bottom index bigger.

First index, m: 4

Second index, n: 3

Try:

Five calculations, each worked by counting the factors of $a$ , with a check using $a = 2$ .

$a^{4} \times a^{3}$ . Four factors and three more make seven: $a^{7}$ . Check: $16 \times 8 = 128 = 2^{7}$ .
$a^{5} \div a^{2}$ . Two pairs cancel and three factors are left: $a^{3}$ . Check: $32 \div 4 = 8 = 2^{3}$ .
$a^{3} \div a^{3}$ . Every factor cancels, so the answer is 1. The law would give $a^{3 - 3} = a^{0}$ . Check: $8 \div 8 = 1$ .
$a^{2} \div a^{5}$ . Three factors are left underneath, so the answer is $\frac{1}{a ^{3}}$ . The law would give an index of $- 3$ . The next lecture shows what that must mean. Check: $4 \div 32 = \frac{1}{8}$ .
$(a^{2})^{3}$ . Three groups of two factors make six: $a^{6}$ . Check: $4^{3} = 64 = 2^{6}$ .

What should you have seen? Multiplying and a power of a power always matched the law. Dividing matched it while the top had more factors. With equal indices, every tile cancelled and the answer was 1. With a bigger bottom index, the leftover factors were underneath. The definition stops at those two cases.

A power of a power: count the groups

What does $(a^{2})^{3}$ mean? The index 3 says: three copies of the thing in the bracket, multiplied together.

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

There are 3 groups, and each group has 2 factors. That is $3 \times 2 = 6$ factors. In general, $(a^{m})^{n}$ is $n$ groups of $m$ factors, so it has $m \times n$ factors. This is the third law.

(a^{m})^{n} = a^{mn}

The standard form lecture said that multiplying the indices is the right rule for a different situation, a power raised to another power. Now you can see why. Adding counts two separate lists of factors. Multiplying counts equal groups.

Stop and think

Are $(2^{3})^{2}$ and $2^{3^{2}}$ the same number?

Show the answer

No. $(2^{3})^{2} = 8^{2} = 64$ , which is $2^{6}$ . In $2^{3^{2}}$ there is no bracket, and the top index is worked out first: $3^{2} = 9$ , so $2^{3^{2}} = 2^{9} = 512$ . The bracket says "make $2^{3}$ first, then square it". Without the bracket, the 2 is raised to the number $3^{2}$ . So the bracket is not decoration. It changes the order, and the order changes the answer.

Why a⁰ has to be 1

This is the hard part of the lecture. Spend time here.

Go back to the hook. The definition counts factors, so it cannot say what $a^{0}$ means. Mathematicians are free to give $a^{0}$ a meaning. But they want one thing very strongly: the laws should keep working. A meaning that breaks the laws would make every calculation with powers unreliable.

So ask the question this way. If the second law is going to work when the two indices are equal, what must $a^{0}$ be?

Take any non-zero number $a$ , and divide $a^{3}$ by itself. There are two ways to work it out.

By the law: $a^{3} \div a^{3} = a^{3 - 3} = a^{0}$ .
By ordinary arithmetic: any non-zero number divided by itself is 1, so $a^{3} \div a^{3} = 1$ .

Both lines are the same calculation, so both answers must be equal. This means that $a^{0} = 1$ . There is no other choice. If $a^{0}$ were anything else, the second law would give the wrong answer here.

That is it. That is the whole idea. $a^{0} = 1$ is not a fact to memorise. It is the only meaning that keeps the laws consistent.

A good argument should agree with every other way of looking at the question, so test it twice more. First, the first law: $a^{0} \times a^{5}$ should be $a^{0 + 5} = a^{5}$ . The only number that leaves $a^{5}$ unchanged when you multiply by it is 1. So the first law agrees.

Second, a pattern. Each time the index goes down by 1, there is one factor fewer, so the value is divided by the base. The standard form lecture used this pattern with tens to reach $1 0^{0} = 1$ . Here it is with twos.

Power	Value	How to get the next value
$2^{4}$	16	divide by 2
$2^{3}$	8	divide by 2
$2^{2}$	4	divide by 2
$2^{1}$	2	divide by 2
$2^{0}$	1

Three different arguments, one answer. So the calculator was right about $5^{0}$ . It gives 1 because that is the only value that fits.

The zero index

For every number $a$ except 0, $a^{0} = 1$ .

This is forced by the law $a^{m} \div a^{n} = a^{m - n}$ with $m = n$ .

Why not $a = 0$ ? Try the argument with 0. It needs $0^{3} \div 0^{3}$ , which is $0 \div 0$ , and that has no meaning. The pattern also fails. One pattern, $0^{3} = 0$ , $0^{2} = 0$ , $0^{1} = 0$ , suggests $0^{0} = 0$ . Another pattern, $3^{0} = 1$ , $2^{0} = 1$ , $1^{0} = 1$ , suggests $0^{0} = 1$ . The two patterns disagree, so no single value keeps everything working. Different parts of mathematics make different choices for $0^{0}$ . At this level, it is simply left without a value.

Stop and think

Is $3 x^{0}$ the same as $(3 x)^{0}$ ? Take $x$ to be any number except 0.

Show the answer

No. In $3 x^{0}$ , the index belongs only to $x$ . So $3 x^{0} = 3 \times x^{0} = 3 \times 1 = 3$ . In $(3 x)^{0}$ , the bracket makes the whole of $3 x$ the base, so $(3 x)^{0} = 1$ . An index belongs to whatever sits directly underneath it: one letter or number, or a whole bracket.

Brackets with more than one factor inside

What happens when the base is a product, as in $(ab)^{3}$ ? Use the definition again.

(ab)^{3} = (ab) \times (ab) \times (ab) = (a \times a \times a) \times (b \times b \times b) = a^{3} b^{3}

The middle step is allowed because the order of multiplication does not matter. So you can collect the $a$ 's together and the $b$ 's together. Each of the 3 copies of the bracket gives one $a$ and one $b$ . So the index goes to every factor inside the bracket. The same counting works for a fraction: $(\frac{a}{b})^{3} = \frac{a \times a \times a}{b \times b \times b} = \frac{a ^{3}}{b ^{3}}$ .

There is one warning. This works because the inside is a product. It does not work for a sum. $(a + b)^{2}$ means $(a + b) \times (a + b)$ , and that is not $a^{2} + b^{2}$ . Test it with numbers: $(3 + 4)^{2} = 49$ , but $3^{2} + 4^{2} = 25$ .

The index laws

For $a$ and $b$ not zero, and whole-number indices $m$ and $n$ :

$a^{m} \times a^{n} = a^{m + n}$ · $a^{m} \div a^{n} = a^{m - n}$ · $(a^{m})^{n} = a^{mn}$

$(ab)^{n} = a^{n} b^{n}$ · $(\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}}$ · $a^{0} = 1$

Using the laws

Worked Example 1 — Multiply, then divide

Simplify $x^{3} \times x^{5} \div x^{4}$ .

Work from left to right. Multiplying adds the indices: $x^{3} \times x^{5} = x^{8}$ . Dividing subtracts them: $x^{8} \div x^{4} = x^{4}$ .

Answer: $x^{4}$ .

Check with $x = 2$ : $8 \times 32 \div 16 = 16$ , and $2^{4} = 16$ . ✓

Worked Example 2 — A bracket with a number inside

Simplify $(2 a^{2} b^{3})^{3}$ .

The bracket is a product of three factors: $2$ , $a^{2}$ and $b^{3}$ . The cube applies to every one of them.

(2 a^{2} b^{3})^{3} = 2^{3} \times (a^{2})^{3} \times (b^{3})^{3} = 8 a^{6} b^{9}

The number 2 is a factor like the others, so it is cubed too: $2^{3} = 8$ .

Check with $a = 2$ and $b = 1$ : the bracket is $2 \times 4 \times 1 = 8$ , and $8^{3} = 512$ . The answer gives $8 \times 64 \times 1 = 512$ . ✓

Worked Example 3 — Two laws in one line

Simplify $(3 x^{2})^{2} \times 2 x^{3}$ .

First the bracket: $(3 x^{2})^{2} = 3^{2} \times (x^{2})^{2} = 9 x^{4}$ .

Then multiply. The numbers multiply, $9 \times 2 = 18$ . The powers of $x$ add their indices, $x^{4} \times x^{3} = x^{7}$ . Answer: $18 x^{7}$ .

Check with $x = 2$ : $(3 \times 4)^{2} \times 2 \times 8 = 144 \times 16 = 2304$ , and $18 \times 128 = 2304$ . ✓

Worked Example 4 — When more factors are underneath

Simplify $(4 a^{3} b)^{2} \div (2 a b^{2})^{2}$ .

Top: $(4 a^{3} b)^{2} = 16 a^{6} b^{2}$ . Bottom: $(2 a b^{2})^{2} = 4 a^{2} b^{4}$ . So the expression is $\frac{16 a ^{6} b ^{2}}{4 a ^{2} b ^{4}}$ .

Deal with each kind of factor separately. Numbers: $16 \div 4 = 4$ . The $a$ 's: six on top and two underneath, so $a^{4}$ is left on top. The $b$ 's: two on top and four underneath. Two pairs cancel, and two $b$ 's are left underneath.

\frac{16 a ^{6} b ^{2}}{4 a ^{2} b ^{4}} = \frac{4 a ^{4}}{b ^{2}}

Check with $a = 2$ and $b = 1$ : $3 2^{2} \div 4^{2} = 1024 \div 16 = 64$ , and $4 \times 16 \div 1 = 64$ . ✓

Worked Example 5 — Changing the base

Write $8^{2} \times 4^{3} \div 2^{5}$ as a single power of 2, and find its value.

The laws need one base. So write every number as a power of 2: $8 = 2^{3}$ and $4 = 2^{2}$ .

(2^{3})^{2} \times (2^{2})^{3} \div 2^{5} = 2^{6} \times 2^{6} \div 2^{5} = 2^{6 + 6 - 5} = 2^{7} = 128

Check: $64 \times 64 = 4096$ , and $4096 \div 32 = 128$ . ✓

One line in the working below breaks a law. Find it.

Spot the error — simplifying (2x³)⁴ ÷ 4x⁵

A student simplifies (2x³)⁴ ÷ 4x⁵ and gets 4x². Tap the line that contains the mistake. Read the explanation for any line you tap.

A student's working for $(2 x^{3})^{4} \div 4 x^{5}$ , with the mistake marked.

$(2 x^{3})^{4} = 2^{4} \times (x^{3})^{4}$ . Correct: the index applies to every factor in the bracket.
$= 16 \times x^{7}$ . The mistake. A power of a power multiplies the indices: $(x^{3})^{4}$ is 4 groups of 3 factors, so it is $x^{12}$ , not $x^{7}$ .
$16 x^{7} \div 4 x^{5} = (16 \div 4) \times x^{7 - 5}$ . This step is done correctly from the line before.
$= 4 x^{2}$ . Correct arithmetic, but the answer is wrong because of line 2.

The right answer is $16 x^{12} \div 4 x^{5} = 4 x^{7}$ . Check with $x = 2$ : $(2 \times 8)^{4} = 65536$ , and $65536 \div 128 = 512 = 4 \times 2^{7}$ .

Three classic mistakes

Trap 1 — Multiplying the bases

$2^{3} \times 2^{4}$ is not $4^{7}$ . The base stays the same, because you are still counting factors of 2: the answer is $2^{7} = 128$ . ( $4^{7}$ is 16 384.) And when the bases are different, as in $2^{3} \times 5^{2}$ , the laws do not combine them at all.

Trap 2 — Forgetting the number in the bracket

$(3 x^{2})^{2}$ is $9 x^{4}$ , not $3 x^{4}$ . The 3 is a factor inside the bracket, so it is squared like every other factor. A quick check with $x = 1$ catches this: $(3)^{2} = 9$ .

Trap 3 — Sharing a power over a sum

$(a + b)^{2}$ is not $a^{2} + b^{2}$ , and $2^{3} + 2^{4}$ is not $2^{7}$ . The laws are about products, never sums. $2^{3} + 2^{4} = 8 + 16 = 24$ , while $2^{7} = 128$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Simplify: (a) $y^{7} \times y^{2} \div y^{6}$ (b) $(p^{4})^{3} \div p^{5}$ (c) $5 x^{0}$ , where $x \neq = 0$ .

Hint 1
Decide which law each step uses: multiplying, dividing, or a power of a power.

Hint 2
In (b), $(p^{4})^{3}$ is 3 groups of 4 factors. In (c), the index 0 belongs only to $x$ .

Full answer
(a) $y^{7 + 2} = y^{9}$ , then $y^{9 - 6} = y^{3}$ . (b) $(p^{4})^{3} = p^{12}$ , then $p^{12 - 5} = p^{7}$ . (c) $5 x^{0} = 5 \times 1 = 5$ . Check (a) with $y = 2$ : $128 \times 4 \div 64 = 8 = 2^{3}$ . ✓
Simplify $(5 p^{3} q)^{2} \times 2 p q^{4}$ .

Hint 1
Square every factor inside the bracket first, including the 5.

Hint 2
$(5 p^{3} q)^{2} = 25 p^{6} q^{2}$ . Now multiply the numbers, then add the indices of $p$ , then of $q$ . Remember that $p$ means $p^{1}$ .

Full answer
$(5 p^{3} q)^{2} = 25 p^{6} q^{2}$ . Then $25 p^{6} q^{2} \times 2 p^{1} q^{4} = 50 p^{7} q^{6}$ . Check with $p = q = 1$ : $25 \times 2 = 50$ . ✓ Check with $p = 2$ , $q = 1$ : $(5 \times 8)^{2} \times 2 \times 2 = 1600 \times 4 = 6400$ , and $50 \times 128 = 6400$ . ✓
Find $n$ in each equation. (a) $3^{n} \times 9^{2} = 3^{10}$ (b) $4^{n} = 8^{4}$

Hint 1
The laws only combine powers with the same base. Write 9 as a power of 3, and write 4 and 8 as powers of 2.

Hint 2
(a) $9^{2} = (3^{2})^{2} = 3^{4}$ . (b) $4^{n} = (2^{2})^{n} = 2^{2 n}$ and $8^{4} = (2^{3})^{4}$ . When two powers of 2 are equal, their indices are equal.

Full answer
(a) $3^{n} \times 3^{4} = 3^{n + 4}$ , so $n + 4 = 10$ and $n = 6$ . Check: $729 \times 81 = 59049 = 3^{10}$ . ✓ (b) $2^{2 n} = 2^{12}$ , so $2 n = 12$ and $n = 6$ . Check: $4^{6} = 4096$ and $8^{4} = 4096$ . ✓
Imagine a sheet of paper 0.1 mm thick that is folded in half again and again. (Real paper cannot be folded nearly this many times.) After $n$ folds there are $2^{n}$ layers. (a) How many times thicker is the pile after 10 folds than after 4 folds? (b) A student says, "20 folds is twice as many as 10 folds, so the pile is twice as thick." Use the laws to show how many times thicker it really is. (c) How thick is the pile after 20 folds, in metres?

Hint 1
"How many times thicker" is a division: $2^{10} \div 2^{4}$ .

Hint 2
For (b), divide $2^{20}$ by $2^{10}$ . For (c), $2^{20} = (2^{10})^{2}$ , and $2^{10} = 1024$ .

Full answer
(a) $2^{10} \div 2^{4} = 2^{6} = 64$ times thicker. (b) $2^{20} \div 2^{10} = 2^{10} = 1024$ times thicker, not 2 times. Doubling the number of folds squares the number of layers, because $2^{20} = (2^{10})^{2}$ . (c) $2^{20} = 102 4^{2} = 1048576$ layers, so the pile is $0.1 \times 1048576 = 104 857.6$ mm, which is about 105 m. Check: 1000 mm make 1 m, and $104 857.6 \div 1000 \approx 104.9$ . ✓

The bigger idea

Look at how this lecture was built. It started with one definition that works for counting numbers. It proved three laws by counting. Then it met a case the definition could not reach, $a^{0}$ , and asked which meaning keeps the laws true. Exactly one meaning did.

This way of extending an idea runs all through mathematics. Negative numbers, fractions and, at Pre-University, powers with any real index were all given meanings so that the old rules keep working. It does not always succeed: $0^{0}$ showed a case where no single value keeps every rule. But when it does succeed, the new meaning is not a choice someone made. It is forced.

The next lecture uses the same question on $a^{2} \div a^{5}$ , the case where the factors ran out on top. It asks what $a^{- 3}$ must mean, and then what $a^{1/2}$ must mean, if $a^{1/2} \times a^{1/2}$ is to equal $a^{1}$ .

Key concepts in this lesson

For a positive whole number $n$ , $a^{n}$ means $n$ factors of $a$ multiplied together, so the index counts factors.
Multiplying powers of the same base adds the indices, because the two lists of factors join into one list.
Dividing powers of the same base subtracts the indices, because each factor underneath cancels one factor on top.
A power of a power multiplies the indices, because it counts equal groups of factors, and an index on a bracket applies to every factor inside it, but never across a sum.
$a^{0} = 1$ for every non-zero $a$ , because it is the only value that keeps the law $a^{m} \div a^{n} = a^{m - n}$ true when $m = n$ .