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ACT Math

ACT Math Help: Radical And Exponential Functions

Review real example questions for Radical And Exponential Functions in ACT Math.

Question 1

The velocity vvv (in ft/sec) after falling ddd feet is estimated by v=64dv = \sqrt{64d}v=64d​. What is the velocity after falling 25 feet?

  1. 10
  2. 40
  3. 89
  4. 1,600
Explanation: This is a formula evaluation question testing radical simplification. Choice B (40) is correct — substitute d = 25: v = √(64 × 25) = √1600 = 40. A faster approach: √(64 × 25) = √64 × √25 = 8 × 5 = 40. Choice A (10) likely comes from computing √64 + √25 = 8 + 5... wait, that's 13. A (10) more likely: student divides instead of multiplying: √(64/25) ≈ 1.6... or computes 64 − 25 = 39, √39 ≈ 6.2. Most likely: student evaluates √64 = 8 and divides by √(25/something). Choice C (89) comes from adding before taking the root: √(64 + 25) = √89 ≈ 9.4, then rounds up to 89 or misreads. Choice D (1,600) correctly multiplies 64 × 25 = 1,600 but forgets to take the square root. Pro tip: When a formula contains a square root, evaluate everything inside the radical first, then take the root. You can also split √(ab) = √a × √b when both are perfect squares — a useful shortcut here since both 64 and 25 are perfect squares.

Question 2

Let P(t)=400(3)t2P(t) = 400(3)^{\frac{t}{2}}P(t)=400(3)2t​. What is the value of P(6)P(6)P(6)?

  1. 2,400
  2. 3,600
  3. 10,800
  4. 32,400
Explanation: This is an exponential models question testing careful evaluation of a fractional exponent. Choice C (10,800) is correct — substitute t = 6: P(6) = 400(3)^(6/2) = 400(3)³ = 400 × 27 = 10,800. Choice A (2,400) treats the exponent as a multiplier: 400 × 6 = 2,400, ignoring the exponential structure entirely. Choice B (3,600) uses exponent 2 instead of 3: 400 × 3² = 400 × 9 = 3,600 — computing 6/2 as 2 rather than 3, or using n − 1 = 2 from sequence thinking. Choice D (32,400) uses exponent 4 instead of 3: 400 × 3⁴ = 400 × 81 = 32,400 — perhaps computing (6/2) + 1 = 4. Pro tip: Always resolve the exponent completely before computing the power. Here, t/2 = 6/2 = 3, so the base 3 is raised to the 3rd power: 3³ = 27. Writing it out as 400 × 3 × 3 × 3 = 400 × 27 avoids confusion about what the exponent is.

Question 3

If x23=16x^{\frac{2}{3}} = 16x32​=16 and x>0x > 0x>0, what is the value of xxx?

  1. 444
  2. 323\frac{32}{3}332​
  3. 646464
  4. 256256256
Explanation: The correct answer is C (64). To solve x^(2/3) = 16, raise both sides to the power 3/2 (the reciprocal of 2/3): x = 16^(3/2). Evaluate: 16^(3/2) = (√16)³ = 4³ = 64. A (4) comes from applying only the square root part: √16 = 4, stopping there without cubing. D (256) results from squaring 16 instead of applying the 3/2 power: 16² = 256. B (32/3) treats the fractional exponent as multiplication: 16 × (2/3) = 32/3. Pro tip: to undo x^(m/n) = k, raise both sides to the power n/m — the reciprocal of the exponent.

Question 4

Given \sqrt6{x} \= y where yyy is a real number, what must be true of xxx?

  1. xxx is an integer.
  2. xxx is a rational number.
  3. xxx is a positive real number.
  4. xxx is a nonnegative real number.
Explanation: This is a domain of expressions question testing the restriction on even-index radicals. Choice D (x is a nonnegative real number) is correct — for ⁶√x to produce a real number y, the radicand x must be ≥ 0. This is because even-index roots of negative numbers are not real. x = 0 gives y = 0, which is valid. So x ≥ 0, meaning x is nonnegative. Choice A (x is an integer) is too restrictive — x = 0.5 gives ⁶√0.5, which is a perfectly valid real number. Choice B (x is rational) is also too restrictive — x = √2 (irrational) gives a valid real 6th root. Choice C (x is a positive real number) excludes x = 0, which produces ⁶√0 = 0, a valid real output. Pro tip: The key constraint for real nth roots: when n is even (2, 4, 6, ...), the radicand must be ≥ 0. When n is odd (3, 5, 7, ...), the radicand can be any real number. Here n = 6 (even), so x ≥ 0. Zero must be included because ⁶√0 = 0 is real and valid.

Question 5

What is 434^343?

  1. 323232
  2. 161616
  3. 646464
  4. 121212
Explanation: To evaluate 434^343, we multiply 4 by itself three times. 43=4×4×4=16×4=644^3 = 4 \times 4 \times 4 = 16 \times 4 = 6443=4×4×4=16×4=64. The exponential notation means repeated multiplication of the base. Choice B (161616) would be 424^242, not 434^343.

Question 6

Which expression equals e−5e^{-5}e−5?

  1. 1/e−51/e^{-5}1/e−5
  2. −e5-e^5−e5
  3. e5e^5e5
  4. 1/e51/e^51/e5
Explanation: The negative exponent rule states that e−5=1/e5e^{-5} = 1/e^5e−5=1/e5. A negative exponent means to take the reciprocal and make the exponent positive. The expression equals 1/e51/e^51/e5. Choice B incorrectly gives −e5-e^5−e5, confusing the negative sign with the negative exponent rule.

Question 7

Simplify: 27\sqrt{27}27​

  1. 54\sqrt{54}54​
  2. 9\sqrt{9}9​
  3. 666
  4. 333\sqrt{3}33​
Explanation: To simplify 27\sqrt{27}27​, we factor out perfect squares from under the radical. 27=9×3=9×3=33\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}27​=9×3​=9​×3​=33​. The simplified form is 333\sqrt{3}33​ since 9 is a perfect square. Choice C (666) would be if we incorrectly added 3+33 + 33+3 instead of multiplying.

Question 8

What is the value of xxx if 2x=322^x=322x=32?

  1. 444
  2. 555
  3. 666
  4. 101010
Explanation: To solve 2^x = 32, we need to express 32 as a power of 2. Since 32 = 2⁵, we have 2^x = 2⁵. When the bases are equal, the exponents must be equal, so x = 5. We can verify: 2⁵ = 32. Choice C incorrectly gives x = 6, which would yield 2⁶ = 64.

Question 9

Simplify: (x3)4(x^3)^4(x3)4

  1. x7x^7x7
  2. x12x^{12}x12
  3. x81x^{81}x81
  4. x16x^{16}x16
Explanation: To simplify (x3)4(x^3)^4(x3)4, we apply the power rule for exponents: (am)n=amn(a^m)^n = a^{mn}(am)n=amn. Therefore, (x3)4=x3×4=x12(x^3)^4 = x^{3 \times 4} = x^{12}(x3)4=x3×4=x12. When raising a power to another power, we multiply the exponents. Choice A incorrectly adds the exponents (3 + 4 = 7) instead of multiplying them.

Question 10

What is the domain of the real-valued function f(x)=x−5f(x) = \sqrt{x - 5}f(x)=x−5​?

  1. x≤5x \le 5x≤5
  2. x≥5x \ge 5x≥5
  3. x>0x > 0x>0
  4. All real numbers
Explanation: The correct answer is B (x ≥ 5). For a square root to produce a real value, the expression under the radical must be non-negative: x − 5 ≥ 0 → x ≥ 5. A (x ≤ 5) correctly identifies 5 as the boundary but flips the direction — thinking the square root limits x to values below 5. C (x > 0) applies a general positivity condition without accounting for the −5 shift in the radicand. D (all real numbers) ignores the square root restriction entirely. Pro tip: set the expression inside the square root greater than or equal to zero, then solve that inequality.