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ACT Math Help: Radical And Exponential Functions

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

Question 1 / 10

0 of 10 answered

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

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Question 1

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

10
40 (correct answer)
89
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 $P(t) = 400(3)^{\frac{t}{2}}$ . What is the value of $P(6)$ ?

2,400
3,600
10,800 (correct answer)
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 $x^{\frac{2}{3}} = 16$ and $x > 0$ , what is the value of $x$ ?

$4$
$\frac{32}{3}$
$64$ (correct answer)
$256$

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 $y$ is a real number, what must be true of $x$ ?

$x$ is an integer.
$x$ is a rational number.
$x$ is a positive real number.
$x$ is a nonnegative real number. (correct answer)

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 $4^3$ ?

$32$
$16$
$64$ (correct answer)
$12$

Explanation: To evaluate $4^3$ , we multiply 4 by itself three times. $4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$ . The exponential notation means repeated multiplication of the base. Choice B ( $16$ ) would be $4^2$ , not $4^3$ .

Question 6

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

$x \le 5$
$x \ge 5$ (correct answer)
$x > 0$
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.

Question 7

Which of the following values is closest to $4 \cdot 10^3 + 3 \cdot 10^{-2}$

40.03
400.03
4,000.03 (correct answer)
40,000.03

Explanation: Scientific notation problems require understanding place value and how powers of 10 work. 4 × 10³ = 4,000 and 3 × 10⁻² = 0.03. Adding these: 4,000 + 0.03 = 4,000.03. The key insight is recognizing that 10³ affects the 4 and 10⁻² affects the 3, but they're in completely different place values so they simply add. Don't let the scientific notation intimidate you—convert to standard form, add, and you're done.

Question 8

Given the function $f(x) = \sqrt[4]{x - 5}$ , what is the domain of $f(x)$ in the real number system?

$x \ge 0$
$x \ge 5$ (correct answer)
$x \le 5$
All real numbers

Explanation: Domain problems for radical functions require understanding that even-indexed roots (square roots, fourth roots, etc.) cannot have negative radicands in the real number system. For f(x) = ⁴√(x - 5), the expression under the radical must be non-negative: x - 5 ≥ 0, so x ≥ 5. Note that even roots allow zero but not negative values (unlike odd roots, which allow any real number). This question specifically states "in the real number system" to clarify that complex numbers are not considered.

Question 9

If $x$ and $y$ are positive integers such that $3^x \cdot 3^y = 81$ , what is the greatest possible value of $x - y$ ?

0
2 (correct answer)
3
4

Explanation: This is an exponents and integer reasoning question. Choice B (2) is correct — since 3^x · 3^y = 3^(x+y), and 81 = 3⁴, we have x + y = 4. Both x and y must be positive integers, so the valid pairs are: (1, 3), (2, 2), and (3, 1). The corresponding values of x − y are −2, 0, and 2. The greatest is 2, achieved when (x, y) = (3, 1). Choice A (0) only considers the symmetric case (x, y) = (2, 2), ignoring the other valid pairs. Choice C (3) would require the pair (3.5, 0.5) or (4, 1) — neither consists of positive integers that sum to 4. Choice D (4) would require the pair (4, 0) — but y must be a positive integer (y > 0), so y = 0 is excluded. Pro tip: When a problem says "positive integers," remember that zero does NOT count. List all valid integer pairs that satisfy the constraint, compute the target expression for each, and identify the maximum.

Question 10

If $\sqrt{2x + 5} = 3$ , what is the value of $x$ ?

2 (correct answer)
4
7
13

Explanation: This is a radical equations question testing the squaring technique. Choice A (2) is correct — square both sides: (√(2x + 5))² = 3² → 2x + 5 = 9. Subtract 5: 2x = 4. Divide by 2: x = 2. Check: √(2(2) + 5) = √9 = 3 ✓. Choice B (4) results from solving 2x = 9 − 1 = 8 → x = 4, subtracting 1 from 9 instead of 5. Choice C (7) comes from adding 5 to 9 instead of subtracting: 2x = 9 + 5 = 14 → x = 7 — flipping the sign when moving 5 to the right side. Choice D (13) results from a double-squaring error: squaring the 3 to get 9, then squaring again before solving, or treating √(2x + 5) = 9 and solving 2x + 5 = 81 → x = 38... Pro tip: After squaring both sides of a radical equation, solve the resulting linear equation normally. Always check your answer by substituting back — squaring can introduce extraneous solutions.