ACT Math study guide · Chapter 4
Exponents and Radicals
Multiplying powers with the same base adds the exponents. Dividing powers with the same base subtracts the exponents. Raising a power to another power multiplies the exponents. These three rules handle the core exponent manipulations used throughout this chapter; negative, zero, and rational exponents add the domain and reciprocal rules introduced below.
A radical simplifies by factoring out the largest perfect square. √50 becomes √25·2 = 5√2, since 25 is the largest perfect-square factor of 50. Two radicals with the same radicand after simplifying can be combined the way like terms are combined.
A rational exponent connects powers and roots directly: when defined in the real numbers, x^(m/n) means the nth root of x raised to the m power. The denominator of the exponent becomes the root; the numerator stays as the power. With an even root, the radicand must be nonnegative for the result to be a real number. A radical equation is solved by isolating the radical and raising both sides to remove it, then checking every candidate solution against the original equation, since that step can introduce an extraneous root.
Where this appears as a model: repeated growth described in words. A bacteria population starting at 100 and doubling every hour follows P(t)=100·2^t, giving P(3)=800 after 3 hours.
▸ Combining and simplifying radicals
Worked Example Simplify √72.
Factor out the largest perfect square.
72 = 36·2
√72 = √36·√2 = 6√2
▸ Rational exponents and rationalizing denominators
A fraction with a radical in the denominator is not considered fully simplified. Rationalizing the denominator multiplies the fraction by a form of 1 built from that radical, which clears it from the denominator without changing the fraction's value.
Worked Example Rationalize the denominator: 3 ⁄ √2.
Multiply by √2/√2.
3√2 ÷ (√2·√2)
3√2 ÷ 2
▸ Practice
Question 1
- AA. x¹⁵
- BB. 2x⁸
- CC. x²
- DD. x⁸
Question 2
- FF. 5√2
- GG. 25√2
- HH. 2√5
- JJ. 10√5
Question 3
- AA. x⁻³
- BB. x³
- CC. x¹⁰
- DD. x⁷
Question 4
- FF. x^(4/3)
- GG. x^(1/4)
- HH. x^(3/4)
- JJ. x⁷
Question 5
- AA. x = 4
- BB. x = 13/2
- CC. x = −4
- DD. x = 5
Question 6
- FF. √7
- GG. √7 ⁄ 7
- HH. 1/7
- JJ. 7√7
Question 7
- AA. x³
- BB. x¹⁰
- CC. x⁷
- DD. x^2.5
Question 8
- FF. 4x⁶
- GG. 2x⁶
- HH. 4x⁵
- JJ. 2x⁵
Question 9
- AA. 2√49
- BB. 7√2
- CC. 14√2
- DD. 49√2
Question 10
- FF. x = 2 only
- GG. x = 5 or x = 2
- HH. x = 5 only
- JJ. no solution
Question 11
- AA. x^(9/25)
- BB. x
- CC. x^(5/2)
- DD. x^(2/5)
Question 12
- FF. 4√5
- GG. √5/4
- HH. 5√5/4
- JJ. 4√5/5
Question 13
- AA. −2
- BB. 2
- CC. −4
- DD. 4
Question 14
- FF. 3√5
- GG. √35
- HH. 2√15
- JJ. 5√3
Question 15
- AA. x⁷
- BB. x²
- CC. x¹²
- DD. x⁹
Question 16
- FF. 5√3
- GG. 3√25
- HH. 15√3
- JJ. 25√3
Question 17
- AA. 6
- BB. 3
- CC. 9
- DD. 81
Question 18
- FF. 6√3
- GG. 2√3
- HH. √3/2
- JJ. 3√3
Question 19
- AA. 2
- BB. 4
- CC. 6
- DD. 8
Question 20
- FF. 7√5
- GG. 5√5
- HH. √65
- JJ. 13√5
▸ Answers and Explanations
Show the answers and explanations
1: D. Same base, add the exponents: x³⁺⁵ = x⁸. Choice B multiplies the result by 2, as if combining like terms instead of applying the exponent rule.
2: F. 50 = 25·2, and √25 = 5, leaving 5√2. Choice G leaves the factor 25 outside the root without taking its square root.
3: B. Same base, add the exponents: −2 + 5 = 3, giving x³. Choice A adds the exponents with a sign error, computing −2−5 instead of −2+5.
4: H. The root (4) becomes the denominator; the power (3) stays the numerator: x^(3/4). Choice F inverts the fraction, placing the root in the numerator and the power in the denominator.
5: D. Squaring gives 2x − 1 = 9, so x = 5; checking, √9 = 3. Valid, no extraneous root here. Choice A subtracts 1 from 9 instead of adding it, giving 2x = 8 and x = 4.
6: G. Multiplying by √7/√7 clears the radical from the denominator: √7/7. Choice F multiplies by √7/√7 but forgets to complete the division by 7 in the denominator.
7: A. Subtract the exponents (same base): 5−2 = 3, giving x³. Choice C adds the exponents instead of subtracting them.
8: F. Square both the coefficient and the exponent: 2²=4 and (x³)²=x⁶, giving 4x⁶. Choice H adds 2 to the exponent instead of multiplying it by 2.
9: B. 98 = 49·2, and √49 = 7, leaving 7√2. Choice D leaves the factor 49 outside the root without taking its square root.
10: H. Squaring gives x²−7x+10=0, so x=5 or x=2. Checking x=5: √4=2, and 5−3=2, matches. Checking x=2: √1=1, but 2−3=−1; 1≠−1, rejected. Only x=5 is valid. Choice G reports both candidate solutions without checking that x = 2 is extraneous.
11: B. Add the exponents: −2/5 + 7/5 = 5/5 = 1, so the result is x. Choice A multiplies the exponents instead of adding them.
12: J. Multiply by √5/√5: 4√5 ÷ 5 = 4√5/5. Choice F multiplies by √5/√5 but forgets to complete the division by 5 in the denominator.
13: A. 1/16 = 4⁻², so x = −2. Choice B drops the negative sign on the exponent.
14: J. √12 = 2√3 and √27 = 3√3; 2√3 + 3√3 = 5√3. Choice F swaps the coefficient and the radicand between the two simplified terms.
15: A. Apply the power rule first: (x⁴)³ = x¹²; then subtract exponents: x¹² ÷ x⁵ = x⁷. Choice C applies the power rule correctly but stops before dividing by x⁵.
16: F. 75 = 25·3, and √25 = 5, leaving 5√3. Choice H multiplies the simplified coefficient by 3 instead of stopping once the perfect square is extracted.
17: C. Raise both sides to the power 2/3: x = 27^(2/3) = (∛27)² = 3² = 9. Choice A takes the cube root of 27, getting 3, but then multiplies by 2 instead of squaring.
18: G. Multiply by √3/√3: 6√3 ÷ 3 = 2√3. Choice F multiplies by √3/√3 but forgets to complete the division by 3.
19: C. 81 = 3⁴, so x−2 = 4, giving x = 6. Choice A subtracts 2 from 4 instead of adding it, mishandling the isolation step.
20: G. √20 = 2√5 and √45 = 3√5; 2√5 + 3√5 = 5√5. Choice H incorrectly adds the values under one radical, computing √(20+45) instead of simplifying each term first.