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Introduction to 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 A small card

A card shows $a^n$ , with $a=\tfrac38$ and $n=4$ . Write the multiplication it represents, without finding the product.
1. Hint 1
  The base supplies the factor, while the exponent counts its copies.
2. Hint 2
  Replace $a$ by $\tfrac38$ and use as many copies as the value of $n$ requires.
Answer

$\tfrac38\times\tfrac38\times\tfrac38\times\tfrac38$ .

Full solution

The base is $\tfrac38$ , and the exponent is $4$ , so four copies of $\tfrac38$ are multiplied.
$\left(\tfrac38\right)^4=\tfrac38\times\tfrac38\times\tfrac38\times\tfrac38$

Answer

$\tfrac38\times\tfrac38\times\tfrac38\times\tfrac38$ .

Key idea

An exponent counts copies of the base in a product.
Problem 1: Mark complete
Problem 2 A calculator record

A calculator record gives $1.5\times1.5=2.25$ . Find $1.5^3$ .
1. Hint 1
  Three factors are needed, and the record already combines two of them.
2. Hint 2
  Multiply the recorded product by one more copy of $1.5$ .
Answer

$3.375$ .

Full solution

The cube uses three factors of $1.5$ , so the recorded square supplies the first two.
$1.5^3=2.25\times1.5$

Multiply to obtain the value.
$2.25\times1.5=3.375$

Answer

$3.375$ .

Key idea

A running product lets you evaluate a power one factor at a time.
Problem 2: Mark complete
Problem 3 A long numeral

A power of ten is written as an ordinary whole number with nine digits. Write it as $10^n$ , giving the value of $n$ .
1. Hint 1
  Count the leading digit separately from the zeros.
2. Hint 2
  The exponent counts the zeros after the leading $1$ .
Answer

$10^8$ , so $n=8$ .

Full solution

A power of ten has a leading $1$ followed by zeros.
Nine digits leave eight places for zeros.
$9-1=8$

The exponent is therefore $8$ .
$10^8=100{,}000{,}000$

Answer

$10^8$ , so $n=8$ .

Key idea

The exponent in a positive whole-number power of ten is one less than its digit count.
Problem 3: Mark complete
Problem 4 A model cube

A model cube has edge length $0.4$ meter. Write the area of one square face and the volume of the cube each as a power of $0.4$ , and find both. (A cube with edge $1$ meter has a volume of $1$ cubic meter.)
1. Hint 1
  A face's area multiplies two of its equal edges, while the cube's volume multiplies three equal edges.
2. Hint 2
  Multiply two copies of $0.4$ for the face area, then one more copy for the volume.
Answer

Face area: $0.4^2=0.16$ square meter. Volume: $0.4^3=0.064$ cubic meter.

Full solution

Each face is a square with side $0.4$ meter, so its area uses two factors of $0.4$ and is read as zero point four squared.
$0.4^2=0.4\times0.4$

Multiply the two factors.
$0.4\times0.4=0.16$

The volume uses three equal edge lengths and is read as zero point four cubed, so multiply the face area by one more edge length.
$0.4^3=0.16\times0.4$

Multiply again.
$0.16\times0.4=0.064$

The face area is $0.16$ square meter, and the volume is $0.064$ cubic meter.

Answer

Face area: $0.4^2=0.16$ square meter. Volume: $0.4^3=0.064$ cubic meter.

Key idea

A cube's face area is its edge squared and its volume is its edge cubed, so the volume is the face area times one more edge.
Problem 4: Mark complete
Problem 5 A mixed expression

Find the value of $(-10)^3\div200+2\times5^2$ .
1. Hint 1
  Finish each power before working on the division or multiplication.
2. Hint 2
  The cube has three negative factors, so its value is negative.
3. Hint 3
  Add the result of the division to the result of the multiplication.
Answer

$45$ .

Full solution

The powers are evaluated first.
$(-10)^3=-1000$ $5^2=25$

Now divide and multiply.
$-1000\div200=-5$ $2\times25=50$

Add these two results.
$-5+50=45$

Answer

$45$ .

Key idea

Powers are evaluated before multiplication and division, which are evaluated before addition.
Problem 5: Mark complete
Problem 6 A forwarded message

In the first round, Lina sends a message to $7$ people. In each later round, every person who received it in the round before forwards it to $7$ new people, and no one receives the message more than once. Write the number of people who receive it in the third round as a power of $7$ and find it. Then find how many people receive it over all three rounds.
1. Hint 1
  The first round reaches $7$ people and each later round multiplies the count by $7$ , so count how many factors of $7$ the third round uses.
2. Hint 2
  Keep a running product: find the second round's count first, then multiply by $7$ once more.
3. Hint 3
  For the total, add the counts of all three rounds, not only the count of the last one.
Answer

$7^3=343$ people in the third round; $399$ people over all three rounds.

Full solution

The first round reaches $7$ people, which is one factor of $7$ .
$7^1=7$

Each of those $7$ people forwards it to $7$ new people, and since no one receives it twice, these are all different people, so the second round uses two factors of $7$ .
$7^2=7\times7$ $7\times7=49$

Each of those $49$ people forwards it to $7$ new people, all different for the same reason, so the third round uses three factors of $7$ .
$7^3=49\times7$ $49\times7=343$

Add the three rounds to find the total.
$7+49+343=399$

So $343$ people receive the message in the third round, and $399$ receive it over all three rounds.

Answer

$7^3=343$ people in the third round; $399$ people over all three rounds.

Key idea

When the first round reaches a number of people and every later round multiplies the count by that same number, each round's count is a power of that number, and the total over several rounds is a sum of those powers.
Problem 6: Mark complete
Problem 7 A written expression

Find the value of $-(-3+1)^4-3\times(-2)^3+(-7)^1$ .
1. Hint 1
  Parentheses determine each base, and any minus sign outside a power waits until that power is evaluated.
2. Hint 2
  The first minus sign sits outside a fourth power, while the cube contains three negative factors.
3. Hint 3
  A first power leaves its base unchanged; finish the multiplication before adding or subtracting.
Answer

$1$ .

Full solution

Inside the first parentheses, $-3+1=-2$ .
The leading minus sign remains outside the power.
$(-2)^4=16$ $-(-2)^4=-16$

Evaluate the cube and first power.
$(-2)^3=-8$ $(-7)^1=-7$

Perform the multiplication before the subtraction.
$3\times(-8)=-24$

Subtracting the negative result adds its opposite.
$-16-(-24)=8$ $8+(-7)=1$

Answer

$1$ .

Key idea

Parentheses determine the base, and the exponent determines how many copies of that base are used.
Problem 7: Mark complete
Problem 8 Mira's size claim

Mira says $10^6$ is twice as large as $10^3$ , because its exponent is twice as large. Is she correct? Explain, and find how many times as large $10^6$ is as $10^3$ .
1. Hint 1
  The exponent counts factors of ten, so write each power of ten as an ordinary number by counting its zeros.
2. Hint 2
  Compare $10^6$ with twice the value of $10^3$ , then divide $10^6$ by $10^3$ to see how many times as large it is.
3. Hint 3
  Dividing a whole number that ends in at least three zeros by $1000$ removes three of those zeros.
Answer

No. $10^6=1{,}000{,}000$ , while twice $10^3$ is only $2000$ ; $10^6$ is $1000$ times as large as $10^3$ .

Full solution

The exponent on ten counts the zeros after the leading $1$ , so $10^3$ is a $1$ followed by three zeros.
$10^3=1000$

In the same way, $10^6$ is a $1$ followed by six zeros.
$10^6=1{,}000{,}000$

Twice $10^3$ is only two thousand.
$2\times1000=2000$

Since $1{,}000{,}000$ is far more than $2000$ , the number $10^6$ is not twice as large as $10^3$ , and Mira is not correct.

To find how many times as large it is, divide.
Dividing by $1000$ removes three of the six zeros.
$1{,}000{,}000\div1000=1000$

So $10^6$ is $1000$ times as large as $10^3$ .

Answer

No. $10^6=1{,}000{,}000$ , while twice $10^3$ is only $2000$ ; $10^6$ is $1000$ times as large as $10^3$ .

Key idea

Doubling the exponent on ten doubles the count of zeros, and each extra zero multiplies the value by ten, so the value grows far more than twice.
Problem 8: Mark complete
Problem 9 Noor's prediction

In $2^2\times3^3$ , each base is replaced by its opposite, written in parentheses so the negative number is the whole base, and the exponents are kept. Noor says the final product changes sign but keeps the same size. Is Noor correct? Explain.
1. Hint 1
  Consider the effect on each power separately before considering their product.
2. Hint 2
  An even number of negative factors has a different sign effect from an odd number.
3. Hint 3
  Compare the signs and sizes of the two factors before and after the replacements.
Answer

Yes. The product changes from $108$ to $-108$ .

Full solution

The original factors have values $2^2=4$ and $3^3=27$ , so
$4\times27=108$

After replacement, the square is unchanged in value, while the cube is negative.
$(-2)^2=4$ $(-3)^3=-27$

The new product is
$4\times(-27)=-108$

The product changes sign and keeps size $108$ , so Noor is correct.

Answer

Yes. The product changes from $108$ to $-108$ .

Key idea

Changing a base to its opposite preserves an even power and changes the sign of an odd power.
Problem 9: Mark complete
Problem 10 Ari's claim

Ari says $1$ is the only positive number equal to its own fourth power. Test $6$ and $0.3$ , then decide whether Ari is right and explain why.
1. Hint 1
  Compare what repeated multiplication does to a base bigger than $1$ and to a positive base less than $1$ .
2. Hint 2
  Each extra factor bigger than $1$ makes a positive product larger, and each extra factor between $0$ and $1$ makes it smaller.
Answer

Yes. $6^4=1296$ and $0.3^4=0.0081$ ; no positive number other than $1$ equals its own fourth power.

Full solution

Test $6$ with a running product of four factors.
$6\times6=36$ $36\times6=216$ $216\times6=1296$
Since $1296$ is greater than $6$ , the number $6$ does not equal its fourth power.

Test $0.3$ the same way.
$0.3\times0.3=0.09$ $0.09\times0.3=0.027$ $0.027\times0.3=0.0081$
Since $0.0081$ is less than $0.3$ , the number $0.3$ does not equal its fourth power either.

For a base greater than $1$ , each extra factor multiplies a positive product by a number bigger than $1$ and makes it larger, so the fourth power is greater than the base.

For a positive base less than $1$ , each extra factor is between $0$ and $1$ and makes the product smaller, so the fourth power is less than the base.

The base $1$ gives $1\times1\times1\times1=1$ , so no positive number other than $1$ equals its own fourth power, and Ari is right.

Answer

Yes. $6^4=1296$ and $0.3^4=0.0081$ ; no positive number other than $1$ equals its own fourth power.

Key idea

Repeated multiplication grows a base above one and shrinks a positive base below one, so no positive number other than one equals its own fourth power.
Problem 10: Mark complete

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