Bluebook SAT Test SAT Exponents and Radicals Practice 4

Question 1

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

Accepted Answer

1

Answer

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.

Question 2

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

Accepted Answer

11/3

Answer

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.

Question 3

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

Accepted Answer

x^(11/6)

Answer

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)).

Question 4

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

Accepted Answer

2

Answer

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.

Question 5

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

Accepted Answer

x = 7 only

Answer

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.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test SAT Exponents and Radicals Practice 4