SAT Exponents and Radicals Practice Flashcards

For x > 0, which of the following is equivalent to (x^(3/2) · x^(−1/4)) / x^(5/4)?

Answer: 1

Combine the exponents in the numerator first: 3/2 + (−1/4) = 6/4 − 1/4 = 5/4. Then subtract the denominator exponent: 5/4 − 5/4 = 0. Since x^0 = 1 for any x > 0, the expression equals 1. A common error is treating the subtraction of the denominator's exponent as addition, which gives x^(10/4), or mis-converting 3/2 to fourths as 5/4 instead of 6/4.

If 2^x · 4^(x+1) = 8^(2x−3), what is the value of x?

Answer: 11/3

Rewrite every term as a power of 2: 4^(x+1) = 2^(2x+2) and 8^(2x−3) = 2^(6x−9). The left side becomes 2^x · 2^(2x+2) = 2^(3x+2). Setting exponents equal: 3x + 2 = 6x − 9 → 11 = 3x → x = 11/3. Choosing 3 is a typical error from stopping after incorrectly computing 3x = 9 instead of carrying the constant correctly.

For x > 0, which of the following is equivalent to √(x³ · ∛(x²))?

Answer: x^(11/6)

Rewrite ∛(x²) = x^(2/3). Inside the square root: x³ · x^(2/3) = x^(9/3 + 2/3) = x^(11/3). Taking the square root gives x^(11/3 · 1/2) = x^(11/6). A common error is computing √(x³) · ∛(x²) separately without first combining inside the radical, which gives x^(3/2) · x^(2/3) = x^(13/6) — but that is the value of √(x³) · ∛(x²), not √(x³ · ∛(x²)).

How many distinct real solutions does (x − 3)^(2/3) = 4 have?

Answer: 2

Rewrite as [(x−3)^(1/3)]² = 4, so (x−3)^(1/3) = ±2. Case 1: (x−3)^(1/3) = 2 → x−3 = 8 → x = 11. Case 2: (x−3)^(1/3) = −2 → x−3 = −8 → x = −5. Checking x = −5: (−8)^(2/3) = [(−8)^(1/3)]² = (−2)² = 4 ✓. Both are valid real solutions. Students often incorrectly assume negative bases under fractional exponents are undefined in the real numbers, or only consider the positive cube root.

Which value(s) of x satisfy √(3x + 4) − √(x + 2) = 2?

Answer: x = 7 only

Isolate one radical: √(3x+4) = 2 + √(x+2). Squaring: 3x+4 = 4 + 4√(x+2) + x+2 → 2x−2 = 4√(x+2) → x−1 = 2√(x+2). Squaring again: x²−2x+1 = 4x+8 → x²−6x−7 = 0 → (x−7)(x+1) = 0, giving x = 7 or x = −1. Checking x = −1: √1 − √1 = 0 ≠ 2 (extraneous). Only x = 7 satisfies the original equation. The trap answer 'both' ignores the mandatory check for extraneous solutions introduced by squaring.

If 5^(2a) = 3, what is the value of 5^(6a) − 5^(4a)?

Answer: 18

Since 5^(2a) = 3, raise both sides to the 3rd power: 5^(6a) = 3³ = 27. Raise both sides to the 2nd power: 5^(4a) = 3² = 9. Therefore 5^(6a) − 5^(4a) = 27 − 9 = 18. This can also be verified by factoring: 5^(4a)(5^(2a) − 1) = 9(3 − 1) = 18. A common error is computing 5^(6a) − 5^(4a) = 5^(2a) = 3 by incorrectly subtracting exponents (which would apply to division, not subtraction).