Fractional Exponents

Nothing in the 3.1 laws ever asked an exponent to be a whole number, so $x^{1 / 2}$ is legal to write, and so far it means nothing. The plan is the one that gave $x^{0}$ and $x^{- 3}$ their meanings in 3.2. Refuse to guess, run the laws on the new symbol, and keep whichever value they force.

Problem

Whatever

2 1^{1 / 2}

turns out to mean, power of a power from 3.1 still applies to it. Apply that law to

{(2 1^{1 / 2})}^{2}

and enter the value.

Show a hint

You do not need to know what $2 1^{1 / 2}$ is. Power of a power multiplies the exponents.
$\frac{1}{2} \cdot 2 = 1$ , and a first power is just the base.

Show the full solution

Power of a power multiplies the exponents, so

{(2 1^{1 / 2})}^{2} = 2 1^{\frac{1}{2} \cdot 2} = 2 1^{1} = 21 .

Whatever number

2 1^{1 / 2}

is, its square has to be

21

, and

21

is not even a perfect square. The law never needed to know the value.

Problem

Here is a second route, independent of the first. The product rule from 3.1 adds exponents, so

16 9^{1 / 2} \cdot 16 9^{1 / 2} = 16 9^{\frac{1}{2} + \frac{1}{2}} = 16 9^{1}

. So

16 9^{1 / 2}

times itself is

169

. Enter the value of

16 9^{1 / 2}

Show a hint

The exponents add to $1$ , so the product is $169$ itself. You are looking for a number that multiplies by itself to make $169$ .
Its square ends in $9$ , so the number has to end in $3$ or $7$ .

Show the full solution

The exponents add to

1

, so

16 9^{1 / 2} \cdot 16 9^{1 / 2} = 169

, and

13 \cdot 13 = 169

, so

16 9^{1 / 2} = 13 .

A different exponent law from the last problem gives the same conclusion.

16 9^{1 / 2}

has to be a number whose square is

169

Problem

The unfinished business first. Two different numbers have a square equal to

121

. Enter the negative one of the two.

Show a hint

Find a number whose square is $121$ first. There is an obvious positive one.
$1 1^{2} = 121$ , and one other number's square is also $121$ . Squaring wipes out a minus sign.

Show the full solution

(- 11)^{2} = (- 11) (- 11) = 121

, so the negative candidate is

- 11

. Both

11

and

- 11

square to

121

, and the next block settles which of the two the symbol

12 1^{1 / 2}

names.

Problem

The last two blocks each ruled out a tempting wrong answer, and this problem offers both of them. Enter the value of

40 0^{1 / 2}

Show a hint

Not half of $400$ . The exponent asks what number squares to $400$ .
Two numbers square to $400$ . By convention $40 0^{1 / 2}$ means the nonnegative one.

Show the full solution

2 0^{2} = 400

, so

40 0^{1 / 2} = 20

. The answer

200

treats the exponent as a factor, and

- 20

also squares to

400

, but by convention

40 0^{1 / 2}

means the nonnegative choice.

Problem

Nothing in the two routes needed the

2

. Run the product rule with three copies,

21 6^{1 / 3} \cdot 21 6^{1 / 3} \cdot 21 6^{1 / 3} = 21 6^{1}

, so

21 6^{1 / 3}

is the number whose cube is

216

. Enter it.

Show a hint

The three exponents add to $1$ , so this is three equal factors multiplying to $216$ .
Try small whole numbers. $5^{3} = 125$ is too small.

Show the full solution

6^{3} = 6 \cdot 6 \cdot 6 = 216

, so

21 6^{1 / 3} = 6

. This time only one number works.

(- 6)^{3} = - 216

, not

216

, so there was no tie between candidates to break.

Problem

The base is negative in both of

(- 343)^{1 / 3}

and

(- 343)^{1 / 2}

, and exactly one of the two names a real number. Enter the value of the one that does.

Show a hint

Ask what sign an even power of a real number can have, and what sign an odd power can have.
No real number squares to a negative. Cubes are different, so look for a number whose cube is $- 343$ .

Show the full solution

(- 7)^{3} = (- 7) (- 7) (- 7) = - 343

, so

(- 343)^{1 / 3} = - 7

. No real number squares to

- 343

, since squares are never negative, so

(- 343)^{1 / 2}

has no value. An odd root of a negative is negative, not missing.

Problem

Power of a power reads

6 4^{2 / 3}

two ways, since

\frac{1}{3} \cdot 2

and

2 \cdot \frac{1}{3}

are the same exponent. So

6 4^{2 / 3}

is both

{(6 4^{1 / 3})}^{2}

and

{(6 4^{2})}^{1 / 3}

. Work out both. They agree, and that value is the answer.

Show a hint

One reading takes the cube root first and then squares. The other squares first and then takes the cube root.
The root-first route is $4^{2}$ . The power-first route has to walk through $6 4^{2} = 4096$ .

Show the full solution

Root first,

6 4^{1 / 3} = 4

and

4^{2} = 16

. Power first,

6 4^{2} = 4096

, and

1 6^{3} = 4096

, so

409 6^{1 / 3} = 16

as well. Both readings give

6 4^{2 / 3} = 16 .

One order stays small the whole way. The other detours through

4096

for the identical answer.

Both orders are forced equal by power of a power, since

\frac{1}{2} \cdot 3 = 3 \cdot \frac{1}{2}

, and both land on

125

. The root-first path never touches a number bigger than

125

, and that is the whole case for taking the root first.

Problem

Evaluate

25 6^{3 / 8}

. The eighth root is invisible until you write the base as a power of something small. Enter the value.

Show a hint

Try writing $256$ as a power of a small number first.
$256 = 2^{8}$ , and then power of a power collapses the whole thing.

Show the full solution

256 = 2^{8}

, so

25 6^{3 / 8} = {(2^{8})}^{3 / 8} = 2^{8 \cdot \frac{3}{8}} = 2^{3} = 8 .

Rewriting the base turned an invisible eighth root into one line of arithmetic.

Problem

A letter base changes none of this. For positive

w

, the product

w^{5 / 6} \cdot w^{13 / 6}

collapses to a single power

w^{k}

. Enter

k

Show a hint

The product rule from 3.1 still adds the exponents, fractions included.
$\frac{5}{6} + \frac{13}{6}$ has a whole-number value.

Show the full solution

The product rule adds the exponents.

\frac{5}{6} + \frac{13}{6} = \frac{18}{6} = 3,

w^{5 / 6} \cdot w^{13 / 6} = w^{3}

and

k = 3

. Multiplying the exponents instead applies power of a power where the product rule belongs.

Problem

All three moves in one problem, the reciprocal, the root and the power. Evaluate

19 6^{- 3 / 2}

. Enter the value as a fraction in lowest terms.

Show a hint

Handle the minus the 3.2 way first. It puts $19 6^{3 / 2}$ under a $1$ .
$19 6^{1 / 2} = 14$ , and then the $3$ cubes it.

Show the full solution

The minus is a reciprocal, so

19 6^{- 3 / 2} = \frac{1}{19 6^{3 / 2}}

. Root first,

19 6^{1 / 2} = 14

, then

1 4^{3} = 2744

, so

19 6^{- 3 / 2} = \frac{1}{2744} .

Reading the minus as a sign on the value gives

- 2744

, wrong in both size and meaning.

Problem

Evaluate

{(\frac{32}{243})}^{2 / 5}

. A power of a fraction raises the top and the bottom separately, the way each base kept its own exponent in 3.1. Enter the value as a fraction in lowest terms.

Show a hint

Take the fifth root of the top and of the bottom first.
$2^{5} = 32$ and $3^{5} = 243$ , and then each root gets squared.

Show the full solution

3 2^{1 / 5} = 2

since

2^{5} = 32

, and

24 3^{1 / 5} = 3

since

3^{5} = 243

. Squaring each root gives

{(\frac{32}{243})}^{2 / 5} = \frac{2^{2}}{3^{2}} = \frac{4}{9} .

Root first kept every number in the work a single digit.

Problem

One equation to close. Find the exponent

k

with

8 1^{k} = 729

. Write both sides as powers of

3

first. Enter

k

as a fraction in lowest terms.

Show a hint

$81$ and $729$ are both powers of $3$ .
$8 1^{k} = {(3^{4})}^{k} = 3^{4 k}$ , and the right side is $3^{6}$ .

Show the full solution

81 = 3^{4}

and

729 = 3^{6}

, so the equation is

3^{4 k} = 3^{6}

, which forces

4 k = 6

and

k = \frac{3}{2} .

Check it root first,

8 1^{3 / 2} = {(8 1^{1 / 2})}^{3} = 9^{3} = 729

. The lesson ends where it began, on an equation only a fractional exponent can answer.

Every root in this lesson was written as an exponent, and every answer came out an integer or a fraction, because every base was a perfect power. Roots come up often enough to have earned a symbol of their own. 3.4 introduces it, along with the question this lesson dodged, what happens when the base is not a perfect power.

Practice these ideas

Practice

You do not need to know what

2 3^{1 / 2}

is to answer this. Enter the value of

(2 3^{1 / 2})^{2}

Show the solution

Power of a power multiplies the exponents, so

(2 3^{1 / 2})^{2} = 2 3^{(1 / 2) \cdot 2} = 2 3^{1} = 23

. The square of

2 3^{1 / 2}

23

by definition, and the law gives that without the value of

2 3^{1 / 2}

ever being needed.

Practice

Evaluate

48 4^{1 / 2}

Show the solution

2 2^{2} = 484

, so

48 4^{1 / 2} = 22

. The exponent asks for the nonnegative number whose square is

484

, and

22

is the one that checks.

Practice

A student reads the

\frac{1}{2}

32 4^{1 / 2}

as a factor and answers

162

. Enter the actual value.

Show the solution

1 8^{2} = 324

, so

32 4^{1 / 2} = 18

. Multiplying by

\frac{1}{2}

gives

162

, but the exponent asks what squares to

324

, and

16 2^{2}

is nowhere near it.

Practice

A student answers

- 31

for

96 1^{1 / 2}

and runs the check

(- 31)^{2} = 961

, which comes out right. The answer is still wrong. Enter the value of

96 1^{1 / 2}

Show the solution

3 1^{2} = 961

, so

96 1^{1 / 2} = 31

. Both

31

and

- 31

square to

961

, and the symbol names the nonnegative one by convention, so no amount of checking by squaring can rescue

- 31

. That candidate is written

- (96 1^{1 / 2})

Practice

Evaluate

685 9^{1 / 3}

Show the solution

1 9^{3} = 6859

, so

685 9^{1 / 3} = 19

. The bounds

1 0^{3} = 1000

and

2 0^{3} = 8000

trap the root in the teens, and only a units digit of

9

cubes to something ending in

9

, so

19

was the only candidate worth testing.

Practice

Evaluate

(- 4913)^{1 / 3}

Show the solution

1 7^{3} = 4913

, so

(- 17)^{3} = - 4913

and

(- 4913)^{1 / 3} = - 17

. Three negative factors leave a negative product, so an odd root of a negative number is negative, not missing.

Practice

(- 361)^{1 / 2}

(- 3375)^{1 / 3}

(- 49)^{1 / 4}

, and

(- 1)^{1 / 5}

, how many name a real number? Enter the count.

Show the solution

The even roots fail, since no real number raised to the power

2

4

is negative. The odd roots are real,

(- 3375)^{1 / 3} = - 15

and

(- 1)^{1 / 5} = - 1

. That makes

2

real values. Parity settles all four before any arithmetic starts.

Practice

For positive

z

z^{3 / 4} \cdot z^{13 / 4} = z^{k}

. Enter

k

Show the solution

The product rule adds the exponents,

\frac{3}{4} + \frac{13}{4} = \frac{16}{4} = 4

. Multiplying them instead is power of a power's move, and that law belongs to a different expression.

Practice

For positive

c

(c^{5 / 12})^{18} = c^{k}

. Enter

k

as a fraction in lowest terms.

Show the solution

Power of a power multiplies the exponents,

\frac{5}{12} \cdot 18 = \frac{90}{12} = \frac{15}{2}

. Nothing requires the exponent to come out whole, so

\frac{15}{2}

is the finished answer.

Practice

For positive

v

\frac{v^{2 / 5}}{v^{22 / 5}} = v^{k}

. Enter

k

Show the solution

The quotient rule subtracts,

\frac{2}{5} - \frac{22}{5} = - \frac{20}{5} = - 4

. Subtracting in the wrong order gives

4

, off by exactly a sign, which is why the rule fixes the order as top minus bottom.

Practice

Evaluate

90 0^{3 / 2}

. Take the root first to keep the numbers small.

Show the solution

90 0^{1 / 2} = 30

and

3 0^{3} = 27000

, so

90 0^{3 / 2} = 27000

. Power first lands in the same place, but it runs

90 0^{3}

, a nine digit number, through the arithmetic on the way.

Practice

Evaluate

{(\frac{576}{841})}^{1 / 2}

. Enter a fraction in lowest terms.

Show the solution

The exponent applies to top and bottom separately,

57 6^{1 / 2} = 24

and

84 1^{1 / 2} = 29

, so the value is

\frac{24}{29}

. It is already in lowest terms, since

29

is prime and does not divide

24

Practice

Enter

67 6^{- 1 / 2}

as a fraction in lowest terms.

Show the solution

The minus flips first,

67 6^{- 1 / 2} = \frac{1}{67 6^{1 / 2}}

, and

2 6^{2} = 676

, so the value is

\frac{1}{26}

. The minus makes a reciprocal, not a negative value, so the answer is a small positive fraction, never

- 26

Practice

Evaluate

102 4^{7 / 10}

Show the solution

1024 = 2^{10}

, so

102 4^{7 / 10} = (2^{10})^{7 / 10} = 2^{7} = 128

. Rewriting the base does all the work. The tenth root of

1024

was never something to hunt for directly.

Practice

Enter

656 1^{- 3 / 8}

as a fraction in lowest terms.

Show the solution

6561 = 3^{8}

, so

656 1^{3 / 8} = (3^{8})^{3 / 8} = 3^{3} = 27

, and the minus flips that to

\frac{1}{27}

. The minus is 3.2's reciprocal move, the fraction is this lesson's root and power, and they stack without interfering.

Practice

Exactly one of

- (78 4^{1 / 2})

and

(- 784)^{1 / 2}

names a real number. Enter its value.

Show the solution

2 8^{2} = 784

, so

78 4^{1 / 2} = 28

and

- (78 4^{1 / 2}) = - 28

. The other expression puts the minus inside the base, and no real number squares to

- 784

, so

(- 784)^{1 / 2}

has no value at all. The parentheses decide which computation you are being asked to run.

Practice

Find the exponent

k

with

10 0^{k} = 100000

. Write both sides as powers of

10

first. Enter

k

as a fraction in lowest terms.

Show the solution

100 = 1 0^{2}

and

100000 = 1 0^{5}

, so the equation reads

(1 0^{2})^{k} = 1 0^{5}

, which is

1 0^{2 k} = 1 0^{5}

. Matching exponents forces

2 k = 5

, so

k = \frac{5}{2}

. Check it root first,

10 0^{5 / 2} = (10 0^{1 / 2})^{5} = 1 0^{5} = 100000

. No whole number works here, which is exactly the kind of equation fractional exponents exist to answer.