The Laws of Exponents with Variables

1.5 built the exponent laws out of counting, and every one of those counts still works when the base is a letter. A letter stands for a number you have not named yet, so $k^{5}$ is five factors of $k$ . What is new is knowing when the laws do not apply. There are two such cases, powers with different bases and terms joined by a plus sign.

Problem

m^{3} \cdot m \cdot m^{4}

is a row of factors of

m

with nothing but multiplication in it. Written as a single power it is

m^{k}

. Count the factors and enter

k

Show a hint

Write out the factors of $m$ in each piece, then count the whole row.
A letter with no exponent written is one factor.

Show the full solution

A letter with no exponent written is one factor, since

m = m^{1}

, so the row is 3 factors then 1 then 4. Joining rows adds the counts, giving

8

. Skipping the bare

m

is the easy miss and gives 7.

Problem

A student writes

n^{2} + n^{5} = n^{7}

, adding the exponents. Test the claim by picking a number, the way 1.4 did. Evaluate

n^{2} + n^{5}

n = 2

and enter the value.

Show a hint

$n^{2}$ and $n^{5}$ are separate terms, so work out each one and then add.
For comparison, $n^{7}$ at $n = 2$ is 128.

Show the full solution

n = 2

n^{2} = 4

and

n^{5} = 32

, so

n^{2} + n^{5} = 4 + 32 = 36

. The claim predicts

n^{7} = 128

, and one mismatch kills it. The product rule applies to a product, and this is a sum.

Problem

Multiply

(4 c^{5}) (7 c^{3})

into a single term. The numbers out front and the powers of

c

do not get the same treatment, so decide what happens to each before you write anything. Enter the result, typed like

6 y^{2}

Show a hint

Gather the numbers together and the $c$ factors together, the 1.3 way.
The 4 and the 7 are factors. The 5 and the 3 are counts.

Show the full solution

Coefficients are factors, so they multiply,

4 \cdot 7 = 28

. Exponents are counts, so they add,

5 + 3 = 8

. The term is

28 c^{8}

. Treating both halves alike gives

28 c^{15}

11 c^{8}

, each applying the wrong operation to half the term.

Problem

\frac{z^{11}}{z^{4}}

, every factor on the bottom cancels a matching factor on top. What remains is a single power

z^{k}

. Enter

k

Show a hint

Four factors cancel four factors. How many of the eleven are left?

Show the full solution

The four factors on the bottom cancel four of the eleven on top, so

11 - 4 = 7

factors survive. The 4 is the count that got cancelled, not the count left over.

Problem

Simplify

\frac{54 k^{10}}{6 k^{3}}

to a single term. The number part and the

k

part are handled separately. Enter the coefficient of the result.

Show a hint

Split it into a number part and a $k$ part.
The coefficients divide. Only the exponents subtract.

Show the full solution

The coefficients divide,

54 \div 6 = 9

, and the exponents subtract,

10 - 3 = 7

, so the term is

9 k^{7}

and the coefficient is

9

. Subtraction belongs to the exponents. Subtracting the numbers out front would give 48.

Problem

(r^{4})^{6}

means six copies of

r^{4}

multiplied together. Written as one power it is

r^{k}

. Count the factors of

r

in that whole product and enter

k

Show a hint

Each copy of $r^{4}$ contributes four factors of $r$ .

Show the full solution

Each of the six copies of

r^{4}

brings four factors of

r

, so the total is

4 \cdot 6 = 24

. The 6 counts copies of a four-factor block, not extra factors, so adding to get 10 uses the product rule where it does not apply.

Both bands start from the same block of two

x

factors. On top a block of three is set beside it and the row simply gets longer, five factors in all, so

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

. Below, that same block of two is copied three times, and three copies of two factors is six factors, so

(x^{2})^{3} = x^{6}

. Side by side adds the counts. Copying multiplies them.

Problem

Both

q^{5} \cdot q^{4}

and

(q^{5})^{4}

collapse to a single power of

q

, and they do not collapse the same way. Write each one as a single power of

q

. Take the exponent on

q

in the collapsed form of

(q^{5})^{4}

and subtract the exponent on

q

in the collapsed form of

q^{5} \cdot q^{4}

. Enter that one number.

Show a hint

One expression sets two rows side by side. The other makes four copies of one row.
$5 + 4$ and $5 \cdot 4$ are not the same number.

Show the full solution

Side by side adds the counts, so

q^{5} \cdot q^{4} = q^{9}

. Four copies multiply them, so

(q^{5})^{4} = q^{20}

. The difference is

20 - 9 = 11

. Reading the two as the same thing gives 0, which is exactly the confusion.

Problem

Expand

(3 b^{7})^{3}

into a single term. The outer exponent makes three copies of everything inside the parentheses, the 3 in front included. Enter the coefficient of the result.

Show a hint

Write the three copies out. What sits inside each one?

Show the full solution

Three copies of

3 b^{7}

hold three 3s and twenty-one

b

s, so the term is

27 b^{21}

and the coefficient is

27

. The outer exponent reaches the coefficient too, so

3 b^{21}

keeps only one of those three 3s.

Problem

Expand

(u^{3} w^{8})^{4}

. Each base carries its own exponent through the expansion. Enter the exponent on

w

in the result.

Show a hint

Four copies of $u^{3} w^{8}$ reorder into all the $u$ factors and then all the $w$ factors.

Show the full solution

Every copy of

u^{3} w^{8}

contains eight

w

factors, so four copies give

u^{12} w^{32}

and the exponent on

w

32

. The outer 4 reaches both bases. Applying it to the first one only leaves

w^{8}

untouched.

Problem

Simplify

(2 y^{4})^{3} \cdot 5 y^{2}

to a single term. Clear the parentheses first, then multiply. Enter the result, typed like

6 k^{5}

Show a hint

The cube reaches the 2 as well as the $y^{4}$ .
Once the parentheses are gone it is one product of two monomials.

Show the full solution

The cube makes three copies of

2 y^{4}

, so the parentheses clear to

8 y^{12}

. Then

8 \cdot 5 = 40

and

12 + 2 = 14

, giving

40 y^{14}

. Skipping the

2^{3}

gives

10 y^{14}

, which is the whole trap.

Problem

Simplify

\frac{(5 f^{4})^{2} \cdot f^{3}}{5 f^{6}}

to a single term, then evaluate that term at

f = 2

. Enter the value, a single number.

Show a hint

Clear the parentheses first, then build one term on top before you divide.
Check what the outside exponent does to the 5, not just to $f^{4}$ .

Show the full solution

(5 f^{4})^{2} = 25 f^{8}

, so the top is

25 f^{8} \cdot f^{3} = 25 f^{11}

. Dividing by

5 f^{6}

gives

5 f^{5}

, and at

f = 2

that is

5 \cdot 32 = 160

. The outside exponent reaches the 5 as well as the

f^{4}

, so

5 f^{8}

is not enough.

Every exponent in this lesson was a positive whole number, and we kept it that way by putting the larger exponent on top of every quotient. 3.2 drops that rule. Once the bottom exponent is allowed to be the larger one, $x^{0}$ and $x^{- 3}$ need a meaning, and the meaning is whatever keeps these laws working.

Practice these ideas

Practice

Write

g^{7} \cdot g^{8}

as a single power

g^{k}

and enter

k

Show the solution

Joining two rows of

g

factors adds the counts,

7 + 8 = 15

. Multiplying the exponents would give

g^{56}

, which counts copies that are not there.

Practice

A student claims

d^{2} + d^{4} = d^{6}

. Test it the 1.4 way. Evaluate

d^{2} + d^{4}

d = 3

and enter the value.

Show the solution

d = 3

d^{2} = 9

and

d^{4} = 81

, so the value is

9 + 81 = 90

. The claim predicts

3^{6} = 729

, so it fails. A plus sign joins terms, not rows of factors.

Practice

A student claims

a^{4} b^{2} = (a b)^{6}

. Test the claim by evaluating

a^{4} b^{2}

a = 2

and

b = 3

, and enter the value.

Show the solution

a = 2

and

b = 3

a^{4} b^{2} = 16 \cdot 9 = 144

. The claim would give

6^{6} = 46656

, so it fails. Adding exponents needs a shared base, and

a

and

b

are different bases.

Practice

Multiply

(7 h^{2}) (2 h^{6})

into a single term and enter its coefficient.

Show the solution

7 \cdot 2 = 14

, so the term is

14 h^{8}

and the coefficient is

14

. The exponents have nothing to do with the coefficient.

Practice

Multiply

(8 p^{3}) (3 p^{4})

into a single term. Enter the result, typed like

5 y^{2}

Show the solution

Coefficients multiply,

8 \cdot 3 = 24

, while exponents add,

3 + 4 = 7

, giving

24 p^{7}

. Adding the coefficients to get

11 p^{7}

is the tempting slip, but they are factors, not terms.

Practice

Simplify

\frac{v^{16}}{v^{6}}

to a single power

v^{k}

and enter

k

Show the solution

Six factors on the bottom cancel six of the sixteen on top, so

16 - 6 = 10

factors of

v

are still standing. Answering 6 reports the count that cancels, not the exponent left behind.

Practice

Simplify

\frac{48 t^{9}}{6 t^{4}}

to a single term. Enter it, typed like

3 y^{2}

Show the solution

The coefficients divide,

48 \div 6 = 8

, and the counts subtract,

9 - 4 = 5

, giving

8 t^{5}

. Carrying the 48 along and only subtracting the exponents leaves the number part untouched.

Practice

(s^{6})^{7}

is seven copies of

s^{6}

multiplied together. Write it as

s^{k}

and enter

k

Show the solution

Each of the seven copies contributes six factors of

s

, so the row holds

6 \cdot 7 = 42

factors. Adding the exponents gives 13, but that is the product rule, and this is a power raised to a power.

Practice

Expand

(5 n^{3})^{3}

into a single term and enter its coefficient.

Show the solution

The outer exponent applies to every factor inside the parentheses, so

(5 n^{3})^{3} = 5^{3} \cdot n^{3 \cdot 3} = 125 n^{9}

and the coefficient is

125

. Leaving the 5 alone gives

5 n^{9}

and drops two of the three 5s.

Practice

Simplify each of

(c^{3})^{8}

and

c^{3} \cdot c^{8}

to a single power of

c

. Enter the exponent of the first minus the exponent of the second.

Show the solution

Stacking eight copies of

c^{3}

multiplies the counts, so

(c^{3})^{8} = c^{24}

. Setting two rows side by side adds them, so

c^{3} \cdot c^{8} = c^{11}

. The difference is

24 - 11 = 13

. Reading both the same way gives 0.

Practice

Expand

(j^{5} z^{3})^{3}

and enter the exponent on

z

Show the solution

Each of the three copies carries three

z

factors, so the result is

j^{15} z^{9}

and the exponent on

z

9

. The 3 outside reaches both letters, not just the first one.

Practice

Simplify

\frac{(2 m^{5})^{4}}{8 m^{11}}

to a single term. Enter it, typed like

7 v^{3}

Show the solution

The parentheses clear to

16 m^{20}

, then

16 \div 8 = 2

and

20 - 11 = 9

, giving

2 m^{9}

. Forgetting

2^{4}

leaves

2 m^{20}

on top and wrecks the coefficient.