Bluebook SAT Test Digital SAT Hard Math Practice 5

Question 1

If x and y are positive real numbers such that log₂(x) + log₄(y) = 5 and log₄(x) + log₂(y) = 7, what is the value of log₂(xy)?

Accepted Answer

8

Answer

Let a = log₂(x) and b = log₂(y). Since log₄(t) = log₂(t)/2, the equations become a + b/2 = 5 and a/2 + b = 7. Adding both equations: (3/2)(a + b) = 12, so a + b = 8. Since log₂(xy) = log₂(x) + log₂(y) = a + b, the answer is 8.

Question 2

A monic cubic polynomial p(x) has roots at x = 1, x = −1, and x = k (where k ≠ ±1). If p(2) = 12, what is the value of k?

Accepted Answer

−2

Answer

Since p(x) is monic with roots 1, −1, and k: p(x) = (x − 1)(x + 1)(x − k) = (x² − 1)(x − k). Substituting x = 2: (4 − 1)(2 − k) = 3(2 − k) = 12, so 2 − k = 4 and k = −2.

Question 3

The quadratic function f(x) = ax² + bx + c has its vertex at (3, −7) and passes through the point (1, 5). What is the value of a + b + c?

Accepted Answer

5

Answer

Key insight: a + b + c = f(1) for any quadratic. Using vertex form f(x) = a(x − 3)² − 7, substitute the point (1, 5): 5 = a(−2)² − 7 = 4a − 7, so a = 3. Then f(1) = 3(1 − 3)² − 7 = 12 − 7 = 5.

Question 4

For what nonzero value of a does the system below have infinitely many solutions?

6x − 4y = 10
9x − 6y = a

Accepted Answer

15

Answer

For infinitely many solutions the equations must be scalar multiples of each other. The ratio of the x-coefficients is 9/6 = 3/2, and −6/−4 = 3/2 confirms the same ratio on y. Multiplying the right side of the first equation by 3/2: (3/2)(10) = 15. Therefore a = 15.

Question 5

A downward-opening parabola has x-intercepts at x = −5 and x = 1 and a maximum value of 18. What is f(0)?

Accepted Answer

10

Answer

The axis of symmetry is x = (−5 + 1)/2 = −2. Write f(x) = a(x + 5)(x − 1). At the vertex x = −2: f(−2) = a(3)(−3) = −9a = 18, so a = −2. Then f(0) = −2(0 + 5)(0 − 1) = −2(5)(−1) = 10.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Digital SAT Hard Math Practice 5