Exponent Rules in Algebra

Multiplication compresses repeated addition into one step. Exponents do the same for repeated multiplication. Every rule in this lesson comes from counting factors. The values of zero, negative, and fractional exponents come from requiring those rules to keep working.

Problem

Ten 2s multiplied together is written

2^{10}

. Multiplying them out one at a time works, but it is slow. Instead split the ten factors into two equal groups of five, compute one group, then finish with a single multiplication. What is

2^{10}

Show a hint

One group of five 2s multiplies out to a number you know well.

Show the full solution

Five 2s multiply to

32

, and the two groups give

32 \cdot 32 = 1024

. That is five multiplications instead of the nine you would do taking one factor at a time.

Problem

The product

8 \cdot 32

is itself a power of 2. Each factor is itself a power of 2, so the whole product must be one too. Write

8 \cdot 32

2^{x}

and enter

x

Show a hint

Count the factors of 2 in each piece before multiplying anything.

Show the full solution

8

is three factors of 2 and

32

is five more, so the product is eight factors of 2 in one row and

x = 8

. Counting factors beats multiplying out to

256

and factoring back down.

Problem

Evaluate

\frac{5^{8}}{5^{5}}

without ever computing

5^{8}

5^{5}

. Write the top and the bottom as strings of factors, cancel the shared ones, then multiply what is left.

Show a hint

How many factors of 5 are left after the cancelling?

Show the full solution

The five factors of 5 on the bottom cancel five of the eight on top, leaving

5 \cdot 5 \cdot 5 = 125

. Brute force gives

390625 \div 3125

and the same answer, but it throws away the structure.

Problem

The expression

(7^{2})^{3}

means three factors of

7^{2}

. Written out, it is a single power of 7, so

(7^{2})^{3} = 7^{x}

for some

x

. Count the total number of 7s and enter

x

Show a hint

Each copy of $7^{2}$ contributes two 7s to the row.

Show the full solution

Three copies of

7^{2}

, each holding two 7s, make

2 \cdot 3 = 6

factors of 7, so

x = 6

. Adding to get

7^{5}

is the common slip, since the 3 counts copies rather than extra factors.

Problem

Walk down the powers of 3.

3^{4} = 81

3^{3} = 27

3^{2} = 9

, and each step down divides by 3. Keep the pattern going past

3^{1}

3^{0}

. What value does the pattern force?

Show a hint

What is 3 divided by 3?

Show the full solution

Each step down divides by 3, so

3^{2} = 9

drops to

3^{1} = 3

, and one more step gives

3 \div 3

. So

3^{0} = 1

. The reflex answer of

0

never fits, since dividing 3 by 3 cannot land there.

Problem

The pattern does not stop at 1. Take two more steps down from

3^{0}

, dividing by 3 at each step, to land on

3^{- 2}

. Enter its value as a fraction.

Show a hint

Each step still divides by 3, even after the values drop below 1.

Show the full solution

One step below

3^{0} = 1

\frac{1}{3}

, and one more is

\frac{1}{9}

, so

3^{- 2} = \frac{1}{9}

. A negative exponent makes a reciprocal, not a negative number, so

- 9

is the wrong turn.

Two rules, one habit of counting. On top, three 6s beside two 6s make five 6s in a row, so

6^{3} \cdot 6^{2} = 6^{5}

, the exponents adding because the factor counts add. Below, each step to the right divides by 10, and keeping that pattern going forces

1 0^{0} = 1

and

1 0^{- 1} = \frac{1}{10}

, the two gold rungs.

Problem

The product rule keeps working when an exponent is negative. Evaluate

2^{- 3} \cdot 2^{7}

by adding the exponents before computing anything.

Show a hint

Add the exponents first. Only then evaluate the single power that remains.

Show the full solution

Add the exponents,

- 3 + 7 = 4

, so the product is

2^{4} = 16

. The long way finds

2^{- 3} = \frac{1}{8}

and

2^{7} = 128

separately and then divides, landing in the same place with more work.

Problem

Suppose the power-of-a-power rule keeps working when an exponent is

\frac{1}{2}

. Then

{(4 9^{1 / 2})}^{2} = 4 9^{\frac{1}{2} \cdot 2} = 4 9^{1} = 49

, so

4 9^{1 / 2}

must be a number whose square is

49

. What positive number is it?

Show a hint

Which positive number, times itself, gives $49$ ?

Show the full solution

7 \cdot 7 = 49

, so

4 9^{1 / 2} = 7

. Both

7

and

- 7

square to

49

, and the next block explains why the notation names the positive one.

Problem

Evaluate

9^{3 / 2}

. The exponent offers two orders, root first or cube first. Both are legal, but only one keeps every number small.

Show a hint

Write $\frac{3}{2}$ as $\frac{1}{2}$ applied first, then $3$ .
What is $9^{1 / 2}$ ?

Show the full solution

Take the root first.

9^{1 / 2} = 3

, and

3^{3} = 27

. Cubing first is legal and gives the same answer, but it routes you through

9^{3} = 729

before the square root, while root first keeps every number small.

Every rule in this lesson came from counting factors. The values of $a^{0}$ , $a^{- n}$ , and $a^{1 / n}$ were then chosen to keep those rules true. Next, 1.6 returns to expressions and evaluates them at given values. Chapter 3 brings variables into the laws you just built.

Practice these ideas

Practice

Warm up with the definition.

5^{4}

means four factors of

5

multiplied together, nothing more. Evaluate it as a single number.

Show the solution

Group the four

5

s into two pairs, each worth

25

, so

5^{4} = 25 \cdot 25 = 625

. Multiplying one factor at a time gets there too, just with more chances to slip.

Practice

The product

121 \cdot 11

is itself a power of

11

, since each of its two factors is one. Write the product as

1 1^{x}

and enter

x

Show the solution

121 = 1 1^{2}

and

11 = 1 1^{1}

, so the product holds

2 + 1 = 3

factors of

11

, and

x = 3

. The slip is entering

1331

, the product itself, when the question asks for the exponent.

Practice

Evaluate

\frac{1 2^{9}}{1 2^{7}}

. Cancel the shared factors before you multiply anything, and only then compute what is left.

Show the solution

The seven factors of

12

on the bottom cancel seven of the nine on top, leaving

1 2^{9 - 7} = 1 2^{2} = 144

. Writing out

1 2^{9}

, a ten-digit number, reaches the same answer the slow way.

Practice

(1 7^{4})^{3}

collapses to a single power of

17

. Count the total factors of

17

it contains and enter the exponent

x

(1 7^{4})^{3} = 1 7^{x}

Show the solution

Each of the three copies of

1 7^{4}

contributes four factors of

17

, so there are

4 \cdot 3 = 12

in all and

x = 12

. Adding to get

7

confuses this with the product rule, where the factors sit side by side instead of in copies.

Practice

Evaluate

1 4^{0} \cdot 1 4^{2}

. One of the two factors is much simpler than it looks, so pin down its value before you multiply.

Show the solution

1 4^{0} = 1

, so the product is

1 \cdot 1 4^{2} = 196

. Any nonzero base to the zero power is

1

, and the reflex answer of

0

is the usual miss.

Practice

Write

5^{- 2}

as a fraction in lowest terms. Decide what the negative exponent actually does to the base before touching any arithmetic.

Show the solution

5^{- 2} = \frac{1}{5^{2}} = \frac{1}{25}

. The minus in the exponent takes a reciprocal and never attaches a sign, so

- 25

and

- \frac{1}{25}

are both wrong turns.

Practice

Evaluate

1 5^{- 4} \cdot 1 5^{6}

. Add the exponents before computing anything, then evaluate the small power that remains.

Show the solution

Add the exponents,

- 4 + 6 = 2

, so the product is

1 5^{2} = 225

. Working out

1 5^{6} = 11390625

and dividing by

1 5^{4}

gets there too, with far more arithmetic.

Practice

Evaluate

(- 1)^{15} + (- 1)^{16}

. Count the negative factors in each power, decide the sign each count forces, then add the two values.

Show the solution

Fifteen negative factors pair off with one left over, so

(- 1)^{15} = - 1

. Sixteen pair off exactly, so

(- 1)^{16} = 1

. The sum is

- 1 + 1 = 0

. An even exponent erases the sign, so a negative base does not always give a negative result.

Practice

Evaluate

16 9^{1 / 2}

. The exponent names the one nonnegative number whose square is

169

, so search for that number directly.

Show the solution

13 \cdot 13 = 169

, so

16 9^{1 / 2} = 13

- 13

squares to

169

as well, but the notation names exactly one number and the convention keeps the nonnegative one.

Practice

Evaluate

(- 27)^{1 / 3}

, the number whose cube is

- 27

. Decide first whether such a number can even exist, then find it.

Show the solution

(- 3)^{3} = - 27

, so

(- 27)^{1 / 3} = - 3

. An odd root of a negative exists because an odd count of negative factors stays negative, while an even root of a negative names no real number.

Practice

Evaluate

25 6^{3 / 4}

. Take the fourth root first and cube the result, keeping every number in the computation small along the way.

Show the solution

25 6^{1 / 4} = 4

since

4^{4} = 256

, and then

4^{3} = 64

. Cubing first is legal but sends you through

25 6^{3} = 16777216

before the fourth root.

Practice

Find the number

x

that makes

4^{x} = 8

true. The powers of

4

you know jump right past

8

, yet an

x

exists. Enter it as a fraction.

Show the solution

Write both sides in base

2

4^{x} = 2^{2 x}

and

8 = 2^{3}

, so

2 x = 3

and

x = \frac{3}{2}

. No whole number works here, since

4^{1} = 4

and

4^{2} = 16

straddle

8