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Laws of Exponents: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Three missing factors

For $h>1$ , complete $h\times\square\times\square\times\square=h^{16}$ by writing the same power of $h$ , with a positive whole-number exponent, in all three boxes.
1. Hint 1
  Count how many copies of $h$ the result needs and how many the known factor supplies; a bare $h$ is one copy.
2. Hint 2
  The three boxes supply the missing copies in three equal shares.
Answer

$h^5$ in each box.

Full solution

The result needs sixteen copies of $h$ and the known factor supplies one, so the boxes supply fifteen copies together.
$16-1=15$

All three boxes hold the same power, so each supplies a third of those copies.
$15\div3=5$

Check the proposed factors.
$h\times h^5\times h^5\times h^5=h^{16}$

Because $h>1$ , powers of $h$ with different exponents are different numbers, so matching the count of copies is the only way.

Answer

$h^5$ in each box.

Key idea

The exponents of the factors in a product of powers of one base add up to the exponent of the result.
Problem 1: Mark complete
Problem 2 A grouped fraction

For numbers $u$ , $v$ , and $w$ with $w\ne0$ , write $\left(\frac{uv}{w}\right)^3$ as a fraction whose numerator is a product of powers of $u$ and $v$ and whose denominator is a power of $w$ .
1. Hint 1
  Each copy of the fraction supplies both numerator factors and a denominator factor.
2. Hint 2
  Put the outer power onto the numerator and onto the denominator first. The numerator is itself a product, so its power reaches each factor.
Answer

$\frac{u^3v^3}{w^3}$ .

Full solution

The outer exponent applies to the entire fraction, so it reaches both the numerator and denominator.
$\left(\frac{uv}{w}\right)^3=\frac{(uv)^3}{w^3}$

The numerator is a product, so each factor receives the exponent.
$(uv)^3=u^3v^3$

The requested form is $\frac{u^3v^3}{w^3}$ .
The denominator is nonzero because $w\ne0$ .

Answer

$\frac{u^3v^3}{w^3}$ .

Key idea

A power of a fraction applies to every factor in its numerator and denominator.
Problem 2: Mark complete
Problem 3 Nested parentheses

Write $\bigl((z^2)^3\bigr)^2$ as a single power of $z$ .
1. Hint 1
  Each outer exponent counts whole groups of the factors already inside.
2. Hint 2
  Combine the inner two exponents, then account for the outermost exponent.
Answer

$z^{12}$ .

Full solution

Three groups of two factors give six factors.
$(z^2)^3=z^6$

The outer square uses that group twice.
$(z^6)^2=z^{12}$

Answer

$z^{12}$ .

Key idea

Nested powers multiply their exponent counts.
Problem 3: Mark complete
Problem 4 Mia's product

For $q\ne0$ , Mia writes the product $q^8\times q^7\times q^4$ and then divides it by $q$ . Write her result as a single power of $q$ .
1. Hint 1
  Each factor in the product adds copies of the same base to the total count.
2. Hint 2
  Total the copies from all three factors first; the division comes last.
3. Hint 3
  The final division removes one copy of $q$ .
Answer

$q^{18}$ .

Full solution

The three factors hold eight, seven, and four copies of $q$ .
$8+7+4=19$

The value before the division is $q^{19}$ .
Divide by the nonzero factor $q$ .
$\frac{q^{19}}{q}=q^{18}$

Answer

$q^{18}$ .

Key idea

Repeated multiplication adds factor counts, and division by a nonzero copy removes one.
Problem 4: Mark complete
Problem 5 A compact product

Write $(x^2y)^4\times(xy^2)^2$ as a product of one power of $x$ and one power of $y$ .
1. Hint 1
  Each outer exponent reaches every factor inside its own parentheses.
2. Hint 2
  After resolving both powers, combine the copies of each base separately.
Answer

$x^{10}y^8$ , or $y^8x^{10}$ .

Full solution

The fourth power reaches both factors in the first parentheses.
$(x^2y)^4=x^8y^4$

The square reaches both factors in the second parentheses.
$(xy^2)^2=x^2y^4$

Combine the powers of $x$ by adding their exponents.
$x^8\times x^2=x^{10}$

Combine the powers of $y$ the same way.
$y^4\times y^4=y^8$

The requested product is $x^{10}y^8$ .
Reversing the order of its two factors gives the same product.

Answer

$x^{10}y^8$ , or $y^8x^{10}$ .

Key idea

An exponent on a product reaches each factor before powers of matching bases are combined.
Problem 5: Mark complete
Problem 6 A paper cutting

A square sheet has side length $3^4$ cm. It is cut into small squares, each with side length $3$ cm, with no waste. How many small squares are made? Give the count as a single power of $3$ .
1. Hint 1
  Find how many small squares fit along one side of the sheet.
2. Hint 2
  The number of rows is the same as the number in each row.
3. Hint 3
  Square the count along one side and simplify the resulting power.
Answer

$3^6$ small squares.

Full solution

Divide the large side length by the small side length to count the squares along one edge.
$\frac{3^4}{3}=3^3$

There are $3^3$ rows with $3^3$ squares in each, so the count is
$(3^3)^2=3^6$

Checking numerically, the large side is $81$ cm and there are $27$ small squares along each edge, giving $729$ squares, which is $3^6$ .

Answer

$3^6$ small squares.

Key idea

Counting equal squares along each edge connects division of powers with squaring a count.
Problem 6: Mark complete
Problem 7 Two fractions together

For $c\ne0$ , write $\frac{c^{17}}{(c^5)^3}\times\frac{(c^7)^2}{c^9}$ as a single power of $c$ .
1. Hint 1
  Each fraction can be simplified before the two results are multiplied.
2. Hint 2
  Each fraction contains one power of a power.
3. Hint 3
  After subtracting the exponent in each denominator, add the two remaining exponent counts.
Answer

$c^7$ .

Full solution

In the first fraction, the denominator is a power of a power.
$(c^5)^3=c^{15}$

The base is nonzero and $17>15$ , so subtract the exponents.
$\frac{c^{17}}{c^{15}}=c^2$

In the second fraction, the numerator is a power of a power.
$(c^7)^2=c^{14}$

Again $14>9$ , so subtract the exponents.
$\frac{c^{14}}{c^9}=c^5$

Combine the two resulting powers by adding their exponents.
$c^2\times c^5=c^7$

Answer

$c^7$ .

Key idea

Simplify grouped powers and quotients before combining their remaining factor counts.
Problem 7: Mark complete
Problem 8 Two expressions in $t$

For every number $t$ , do $t^3\times t^5$ and $(t^2)^4$ give the same value? Explain without choosing a particular value of $t$ .
1. Hint 1
  Both expressions can be understood by counting copies of $t$ .
2. Hint 2
  One joins two groups, while the other repeats a group four times.
Answer

Yes. Both equal $t^8$ .

Full solution

The first expression joins three copies of $t$ to five copies.
$t^3\times t^5=t^8$

The second uses four groups of two copies.
$(t^2)^4=t^8$

Both are products of eight copies of the same number, so they agree for every $t$ , including zero.

Answer

Yes. Both equal $t^8$ .

Key idea

Different groupings give equal powers when they contain the same total number of factors.
Problem 8: Mark complete
Problem 9 Jo's shortcut

A calculator finds the fifth power of its input and subtracts the cube of the input. Jo says the result is the square of the input for every number. Is Jo correct? Explain.
1. Hint 1
  Jo's claim subtracts the exponents. Recall which operation on powers of the same base that rule belongs to.
2. Hint 2
  At the input $2$ , work out the fifth power and the cube as plain numbers, subtract, and compare with the square of $2$ .
Answer

No. For example, at input $2$ the calculator gives $24$ , while $2^2=4$ .

Full solution

One input that fails is enough; try $2$ .
Its fifth power is $2^5=32$ and its cube is $2^3=8$ .
The calculator subtracts them.
$32-8=24$

The claimed square is
$2^2=4$

The values differ, so this input disproves the claim.
Subtracting exponents is the rule for dividing powers of the same base, as in $\frac{2^5}{2^3}=2^2$ , not for subtracting their values.

Answer

No. For example, at input $2$ the calculator gives $24$ , while $2^2=4$ .

Key idea

Subtracting exponents is the rule for dividing powers of the same nonzero base, not for subtracting their values; no exponent law turns a difference of powers into one power.
Problem 9: Mark complete
Problem 10 Pat's two equalities

Pat checked $\left(\frac{15}{5}\right)^n=\frac{15^n}{5^n}$ and $(15-5)^n=15^n-5^n$ at $n=1$ , and both worked. Pat concludes that both work for every positive whole-number exponent. Test $n=2$ and explain whether the conclusion follows.
1. Hint 1
  An equality working for one exponent need not work for another.
2. Hint 2
  Evaluate the grouped base before squaring on each left side, then evaluate the corresponding right side.
Answer

No. At $n=2$ , the quotient equality holds, with $9$ on both sides, but the difference equality fails, with $100$ on the left and $200$ on the right.

Full solution

For the quotient equality, evaluate the quotient before squaring on the left.
$\left(\frac{15}{5}\right)^2=9$

The right side divides the separate squares, $225$ by $25$ .
$\frac{15^2}{5^2}=9$

For the difference equality, evaluate the difference before squaring on the left.
$(15-5)^2=100$

The right side subtracts the separate squares, $25$ from $225$ .
$15^2-5^2=200$

The power-of-a-quotient law proved in the lesson makes the first equality hold for every positive whole-number exponent, but this check shows that the exponent does not spread across the difference.
Working at $n=1$ did not establish the claim for all exponents.

Answer

No. At $n=2$ , the quotient equality holds, with $9$ on both sides, but the difference equality fails, with $100$ on the left and $200$ on the right.

Key idea

Checking one exponent does not establish a rule for every exponent.
Problem 10: Mark complete

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