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.

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.

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 $4^3$.

The negative exponent rule states that $e^{-5} = 1/e^5$. A negative exponent means to take the reciprocal and make the exponent positive. The expression equals $1/e^5$. Choice B incorrectly gives $-e^5$, confusing the negative sign with the negative exponent rule.

To simplify $\sqrt{27}$, we factor out perfect squares from under the radical. $\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$. The simplified form is $3\sqrt{3}$ since 9 is a perfect square. Choice C ($6$) would be if we incorrectly added $3 + 3$ instead of multiplying.

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.

To simplify $(x^3)^4$, we apply the power rule for exponents: $(a^m)^n = a^{mn}$. Therefore, $(x^3)^4 = x^{3 \times 4} = x^{12}$. When raising a power to another power, we multiply the exponents. Choice A incorrectly adds the exponents (3 + 4 = 7) instead of multiplying them.

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.

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.

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.