Rational Exponents
An exponent can be a fraction. The bottom number is a root. The top number is a power.
What you'll be able to do
- Rewrite a root as a power, and a power as a root.
- Find values like 32³ᐟ⁵ and 9⁻³ᐟ² without a calculator.
- Simplify letter expressions with fraction exponents.
- Solve equations like 2x³ᐟ² + 5 = 59.
The big idea
The bottom number is a root. 64¹ᐟ³ is the cube root of 64. It is 4, because 4³ = 64.
The top number is a power. 1,000²ᐟ³ means take the cube root, then square it. The cube root of 1,000 is 10, and 10² = 100.
Root first. You can do the root or the power first. The root first keeps the numbers small.
A minus sign means flip. 9⁻³ᐟ² = 1 ÷ 9³ᐟ². Since 9³ᐟ² = 3³ = 27, the value is 1/27. It is not negative.
The old rules still work. To multiply, add the exponents: x¹ᐟ² · x¹ᐟ⁴ = x³ᐟ⁴. For a power of a power, multiply them: (x⁶)²ᐟ³ = x⁴.
Undo a power with its flip. To solve x³ᐟ² = 1,000, raise both sides to the 2/3 power. Then x = 1,000²ᐟ³ = 100.
Odd roots of negatives are fine. (−27)¹ᐟ³ = −3, because (−3)³ = −27. But (−9)¹ᐟ² has no real value. No real number squared is negative.
Words you need
- Exponent
- The small raised number. In 32³ᐟ⁵, the exponent is 3/5.
- Base
- The number being raised to a power. In 32³ᐟ⁵, the base is 32.
- Power
- A base with an exponent, like 2⁵. It also means the exponent itself.
- Rational exponent
- An exponent that is a fraction, like 2/3.
- Square root
- A number that gives the first number when you multiply it by itself. √25 = 5.
- Cube root
- A number you use 3 times as a factor to get the first number. ∛8 = 2.
- Root
- ⁿ√a is a number you use n times as a factor to get a. ⁴√81 = 3.
- Index
- The small number on a root sign. In ⁴√81, the index is 4.
- Radical
- A root sign and what is under it, like √5 or ∛x.
- Factor
- A number you multiply by. In 2 · 2 · 2, 2 is a factor.
- Product
- The answer when you multiply.
- Reciprocal
- 1 divided by a number. The reciprocal of 8 is 1/8.
- Variable
- A letter that stands for a number, like x.
- Expression
- Numbers and letters joined by +, −, × or ÷. It has no = sign.
- Simplify
- Write it in a shorter form with the same value.
- Coefficient
- The number in front of a letter. In 70m, it is 70.
- Constant
- A number that does not change, like 70 in E = 70m³ᐟ⁴.
- Logarithm
- The power a base needs to make a number. log₂(8) = 3, because 2³ = 8.
Exponent form and root form
| Exponent form | Root form | Value |
|---|---|---|
| 16¹ᐟ² | √16 | 4 |
| 64¹ᐟ³ | ∛64 | 4 |
| 81¹ᐟ⁴ | ⁴√81 | 3 |
| 1,000²ᐟ³ | (∛1,000)² | 100 |
| 9⁻¹ᐟ² | 1 ÷ √9 | 1/3 |
Worked examples
Example 1
Find 32³ᐟ⁵ and 243⁻²ᐟ⁵ without a calculator.
- The 5 on the bottom means the 5th root. ⁵√32 = 2, because 2⁵ = 32.
- The 3 on top means cube it. 2³ = 8. So 32³ᐟ⁵ = 8.
- For 243⁻²ᐟ⁵, flip first. It is 1 ÷ 243²ᐟ⁵.
- ⁵√243 = 3, and 3² = 9. So 243⁻²ᐟ⁵ = 1/9.
Answer: 32³ᐟ⁵ = 8 and 243⁻²ᐟ⁵ = 1/9
Example 2
Simplify (64x⁶y⁻³)²ᐟ³. Write the answer with positive exponents. All variables are positive.
- Raise each part to the 2/3 power.
- 64²ᐟ³ = (∛64)² = 4² = 16.
- Multiply the exponents: (x⁶)²ᐟ³ = x⁴ and (y⁻³)²ᐟ³ = y⁻².
- Move y⁻² to the bottom to make its exponent positive. The answer is 16x⁴/y².
Answer: 16x⁴/y²
Example 3
Solve 2x³ᐟ² + 5 = 59.
- Subtract 5 from both sides: 2x³ᐟ² = 54.
- Divide both sides by 2: x³ᐟ² = 27.
- Raise both sides to the flip power, 2/3: x = 27²ᐟ³.
- 27²ᐟ³ = (∛27)² = 3² = 9. Check: 9³ᐟ² = 3³ = 27.
Answer: x = 9
Common mistakes
Multiplying by the exponent. 64²ᐟ³ is not 64 × 2/3. Take the cube root, then square it: 64²ᐟ³ = 16.
Making the answer negative. A minus sign in the exponent means flip. 81⁻¹ᐟ⁴ = 1/3, not −3.
Flipping the fraction. The root goes on the bottom. ∛(x⁴) = x⁴ᐟ³, not x³ᐟ⁴.
Adding exponents for a power of a power. (x²ᐟ³)³ = x², not x¹¹ᐟ³. Multiply the exponents.
Practice: 20 questions
11 of 20 are word problems. For a word problem, sketch it first and mark what you know.
Tier 1 Foundations Use the rule directly
- Rewrite ⁵√(x³) and 1 / ⁴√(y⁷) with fractional exponents, and rewrite a³ᐟ² and (8m)¹ᐟ³ in radical form. Simplify any numerical roots. All variables are positive.
- A cube-shaped shipping crate has a volume of 8 cubic feet. The area A of one square face of a cube with volume V is A = V²ᐟ³. Tenzin calculates the area of one face as 8²ᐟ³ = 16/3 square feet. Find his mistake and give the correct area in square feet. Word problem
- Without a calculator, find the exact value of 27²ᐟ³ and the exact value of 16³ᐟ⁴. Which expression is greater, and by how much?
- Simplify a¹ᐟ² · a¹ᐟ³ and b⁷ᐟ⁴ ÷ b¹ᐟ⁴. Write your answers with positive exponents. All variables are positive.
- Biologists model a mammal's resting energy use with the formula E = 70m³ᐟ⁴, where m is the animal's mass in kilograms and E is the energy used in kilocalories per day. Kwesi weighs a dog whose mass is 16 kilograms. How many kilocalories per day does the dog use at rest? Word problem
Tier 2 Intermediate One more step to think through
- A mammal's resting heart rate R, in beats per minute, is modeled by R = 240m⁻¹ᐟ⁴, where m is the animal's mass in kilograms. Marisol checks a horse with a mass of 625 kilograms. What is the horse's predicted resting heart rate in beats per minute? Word problem
- Simplify (16x⁸y⁻⁴)³ᐟ⁴. Write your answer with positive exponents only. All variables are positive.
- Nikolai simplifies 8⁻¹ᐟ³ · (x¹ᐟ²)³ and writes −2x⁷ᐟ². He made one mistake on the number and one mistake on the variable. Find both mistakes and write the correct simplified expression with positive exponents. Assume x is positive.
- For a planet orbiting a star at an average distance of d astronomical units, the length of one orbital year is T = d³ᐟ² Earth years. Planet A orbits at an average distance of d = 4 astronomical units. Planet B orbits at d = 9 astronomical units. Zayd compares their orbits. Which planet has the longer orbital year, and by how many Earth years? Word problem
- Solve the equation 2x³ᐟ⁴ − 5 = 11 for x.
Tier 3 Advanced Several steps - find the hidden structure
- A mammal's resting energy use E, in kilocalories per day, is modeled by E = 70m³ᐟ⁴, where m is the animal's mass in kilograms. A zoo animal uses 1,890 kilocalories per day at rest. What is the animal's mass in kilograms? Word problem
- Haruki orders 32⁻²ᐟ⁵, 27⁻¹ᐟ³, and 125⁻¹ᐟ³ from least to greatest without a calculator. Find the exact fraction for each expression and write the three original expressions in order from least to greatest.
- The total surface area S, in square centimeters, of a cube with volume V cubic centimeters is S = 6V²ᐟ³. Astrid molds a clay cube with a volume of 125 cubic centimeters. She then molds a second clay cube whose total surface area is 384 square centimeters. What is the total surface area of the first cube in square centimeters, and what is the volume of the second cube in cubic centimeters? Word problem
- Simplify (27x⁹ᐟ⁴y⁻³)¹ᐟ³ ÷ (x¹ᐟ⁴y⁻²). Write your answer with positive exponents only. All variables are positive.
- Pranav types (−8)¹ᐟ³ and (−4)¹ᐟ² into his graphing calculator. It shows −2 for the first one and an error for the second one. Pranav says the calculator is wrong about the first one, because a negative number can't have a fractional power. Find his mistake. Explain why the first has a real value and the second has none. Word problem
Tier 4 Challenge Stretch problems, some above grade level
- Define x²ᐟ³ as (∛x)². When solving the equation x²ᐟ³ = 9, a student raises both sides to the 3/2 power, gets x = 9³ᐟ² = 27, and says 27 is the only real solution. Find what the student missed and give all real solutions to the equation.
- Biologists use E = 70m³ᐟ⁴ to model a mammal's daily resting energy E, in kilocalories, from its mass m in kilograms. A capybara has 16 times the mass of a guinea pig. Chiamaka compares both their total daily energy use and their energy use per kilogram of body mass, which is E ÷ m. How many times as much total daily energy does the capybara use as the guinea pig? Which animal uses more energy per kilogram of its own body mass, and how many times as much? Word problem
- Mateo writes the orbit rule for planets as T = d⁹ᐟ⁴ ÷ ⁴√(d³). Here d is a planet's average distance from the Sun in astronomical units, and T is its year in Earth years. Simplify his rule to a single fractional power of d. A planet orbits at d = 9. How many times as long as Earth's year (d = 1) is its year? Word problem
- A microbiologist studies a cell culture across two growth phases. During Phase 1, the number of cells is multiplied by 8¹ᐟ². During Phase 2, the number of cells is multiplied by 4³ᐟ⁴. Over both phases combined, the starting population is multiplied by 2ᵏ. Find the exact value of k, and state how many times as large the cell population is after both phases together. Word problem
- A logarithm tells you a power. log₁₀(1,000) = 3 because 10³ = 1,000. Sound is measured in decibels. A sound that is I times as strong as the softest sound you can hear has a loudness of 10 · log₁₀(I) decibels. A fan is 10,000³ᐟ⁴ times as strong as the softest sound. A street is 100,000⁴ᐟ⁵ times as strong. How loud is each sound in decibels? How many decibels louder is the street? Word problem Above grade level
Answers
Tier 1 Foundations
- x³ᐟ⁵, y⁻⁷ᐟ⁴ or 1 / y⁷ᐟ⁴, √(a³) or (√a)³, and 2∛m. The index of the radical is the denominator of the exponent, so ⁵√(x³) = x³ᐟ⁵ and 1 / ⁴√(y⁷) = 1 / y⁷ᐟ⁴ = y⁻⁷ᐟ⁴. For a³ᐟ², the denominator 2 means a square root and the numerator 3 is the power, giving √(a³). For (8m)¹ᐟ³, take the cube root of the whole product: ∛(8m) = 2∛m.
- Tenzin multiplied the base 8 by the exponent 2/3 instead of taking the cube root of 8 and squaring the result; the correct area is 4 square feet. The exponent 2/3 means take the cube root first and then raise the result to the 2nd power, not multiply by 2/3. Find the cube root of 8: 8¹ᐟ³ = ∛8 = 2 feet for the length of one edge. Square that edge length: 2² = 4 square feet.
- 27²ᐟ³ = 9 and 16³ᐟ⁴ = 8; 27²ᐟ³ is greater by 1. Rewrite 27²ᐟ³ as (∛27)² = 3² = 9. Rewrite 16³ᐟ⁴ as (⁴√16)³ = 2³ = 8. Compare the results: 9 − 8 = 1, so 27²ᐟ³ is greater by 1.
- a⁵ᐟ⁶ and b³ᐟ². When multiplying with the same base, add the exponents: 1/2 + 1/3 = 3/6 + 2/6 = 5/6, giving a⁵ᐟ⁶. When dividing with the same base, subtract the exponents: 7/4 − 1/4 = 6/4 = 3/2, giving b³ᐟ².
- 560 kilocalories per day. Substitute m = 16 into the formula: E = 70 · 16³ᐟ⁴. Evaluate 16³ᐟ⁴ by taking the fourth root of 16 and cubing it: (⁴√16)³ = 2³ = 8. Multiply by 70: 70 × 8 = 560 kilocalories per day.
Tier 2 Intermediate
- 48 beats per minute. Substitute m = 625 into the formula: R = 240 · 625⁻¹ᐟ⁴. A negative exponent means the reciprocal, so 625⁻¹ᐟ⁴ = 1 / ⁴√625 = 1/5. Multiply by 240: 240 × (1/5) = 48 beats per minute.
- 8x⁶ / y³. Raise the coefficient 16 to the 3/4 power: 16³ᐟ⁴ = (⁴√16)³ = 2³ = 8. Multiply the exponent of each variable by 3/4: (x⁸)³ᐟ⁴ = x⁶ and (y⁻⁴)³ᐟ⁴ = y⁻³. Rewrite y⁻³ in the denominator so all exponents are positive: 8x⁶ / y³.
- 8⁻¹ᐟ³ is 1/2, not −2, and (x¹ᐟ²)³ is x³ᐟ², not x⁷ᐟ²; the correct expression is x³ᐟ² / 2. A negative exponent takes the reciprocal rather than making the number negative, so 8⁻¹ᐟ³ = 1 / ∛8 = 1/2. Raising a power to a power multiplies the exponents rather than adding them, so (x¹ᐟ²)³ has exponent 1/2 × 3 = 3/2, giving x³ᐟ². Combine the two parts: (1/2)x³ᐟ², or x³ᐟ² / 2.
- Planet B has the longer orbital year, by 19 Earth years. Find Planet A's orbital year: T = 4³ᐟ² = (√4)³ = 2³ = 8 Earth years. Find Planet B's orbital year: T = 9³ᐟ² = (√9)³ = 3³ = 27 Earth years. Subtract: 27 − 8 = 19, so Planet B's orbital year is 19 Earth years longer.
- x = 16. Add 5 to both sides and divide by 2 to isolate the power: 2x³ᐟ⁴ = 16, so x³ᐟ⁴ = 8. Raise both sides to the reciprocal power 4/3: x = 8⁴ᐟ³. Evaluate 8⁴ᐟ³: (∛8)⁴ = 2⁴ = 16.
Tier 3 Advanced
- 81 kilograms. Substitute E = 1,890 into the formula: 70m³ᐟ⁴ = 1,890. Divide both sides by 70: m³ᐟ⁴ = 27. Raise both sides to the 4/3 power: m = 27⁴ᐟ³ = (∛27)⁴ = 3⁴ = 81 kilograms.
- 125⁻¹ᐟ³ = 1/5, 32⁻²ᐟ⁵ = 1/4, and 27⁻¹ᐟ³ = 1/3; from least to greatest: 125⁻¹ᐟ³, 32⁻²ᐟ⁵, 27⁻¹ᐟ³. Evaluate each power as a positive unit fraction: 32⁻²ᐟ⁵ = 1 / (⁵√32)² = 1 / 2² = 1/4. Evaluate the cube roots: 27⁻¹ᐟ³ = 1 / ∛27 = 1/3, and 125⁻¹ᐟ³ = 1 / ∛125 = 1/5. Since 1/5 < 1/4 < 1/3, the order from least to greatest is 125⁻¹ᐟ³, 32⁻²ᐟ⁵, 27⁻¹ᐟ³.
- The first cube has a total surface area of 150 square centimeters; the second cube has a volume of 512 cubic centimeters. For the first cube, substitute V = 125: S = 6 · 125²ᐟ³ = 6 · (∛125)² = 6 · 25 = 150 square centimeters. For the second cube, set 6V²ᐟ³ = 384 and divide by 6: V²ᐟ³ = 64. Raise both sides to the 3/2 power: V = 64³ᐟ² = (√64)³ = 8³ = 512 cubic centimeters.
- 3x¹ᐟ²y. Apply the outside 1/3 exponent to each factor in the first group: 27¹ᐟ³ = 3, (x⁹ᐟ⁴)¹ᐟ³ = x³ᐟ⁴, and (y⁻³)¹ᐟ³ = y⁻¹. Divide by x¹ᐟ⁴y⁻² by subtracting the exponents of like bases: for x, 3/4 − 1/4 = 2/4 = 1/2; for y, −1 − (−2) = 1. Combine the simplified factors: 3x¹ᐟ²y.
- (−8)¹ᐟ³ equals the real number −2 because (−2)³ = −8, whereas (−4)¹ᐟ² has no real value because no real number multiplied by itself equals a negative number. Rewrite (−8)¹ᐟ³ as the cube root ∛(−8). Because an odd product of negative numbers is negative, (−2) × (−2) × (−2) = −8, so (−8)¹ᐟ³ = −2. Rewrite (−4)¹ᐟ² as the square root √(−4). Any real number squared is 0 or positive, so no real number squares to −4.
Tier 4 Challenge
- The student missed the negative solution x = −27; the two real solutions are x = 27 and x = −27. Write the equation as (∛x)² = 9. Because squaring either 3 or −3 gives 9, the cube root ∛x can equal 3 or −3. Cube both values: x = 3³ = 27 or x = (−3)³ = −27. Both check because (∛(−27))² = (−3)² = 9.
- The capybara uses 8 times as much total daily energy; the guinea pig uses 2 times as much energy per kilogram of body mass as the capybara. Multiplying mass m by 16 multiplies total energy E by 16³ᐟ⁴ = (⁴√16)³ = 2³ = 8. Energy per kilogram is E ÷ m = 70m³ᐟ⁴ ÷ m¹ = 70m⁻¹ᐟ⁴. Multiplying mass by 16 multiplies energy per kilogram by 16⁻¹ᐟ⁴ = 1/2, so the capybara uses half as much energy per kilogram, meaning the guinea pig uses 2 times as much per kilogram.
- T = d³ᐟ²; the year is 27 times as long. Rewrite the radical as a fractional exponent: ⁴√(d³) = d³ᐟ⁴. Subtract the exponents to divide like bases: 9/4 − 3/4 = 6/4 = 3/2, so T = d³ᐟ². Compare d = 9 with d = 1: 9³ᐟ² ÷ 1³ᐟ² = (√9)³ = 3³ = 27 times as long.
- k = 3; the population is 8 times as large. Rewrite each factor with base 2: 8¹ᐟ² = (2³)¹ᐟ² = 2³ᐟ², and 4³ᐟ⁴ = (2²)³ᐟ⁴ = 2⁶ᐟ⁴ = 2³ᐟ². Multiply the two growth factors by adding their exponents: 2³ᐟ² · 2³ᐟ² = 2³ᐟ²⁺³ᐟ² = 2³. Thus k = 3, and the population is multiplied by 2³ = 8.
- The fan is 30 decibels and the street is 40 decibels; the street is 10 decibels louder. 10,000³ᐟ⁴ = (⁴√10,000)³ = 10³ = 1,000, so the fan is 10 · log₁₀(1,000) = 10 · 3 = 30 decibels. 100,000⁴ᐟ⁵ = (⁵√100,000)⁴ = 10⁴ = 10,000, so the street is 10 · log₁₀(10,000) = 10 · 4 = 40 decibels. The street is 40 − 30 = 10 decibels louder.
Stuck on this topic?
Trouble here often traces back to an earlier skill. Our free check-in finds the exact gap. A tutor then works on it with your child.
Find gaps free More practice