NBT Mathematics: Exponents, Surds and Equations Flashcards

Simplify: (9^(x+1) − 9^x) / (3^(2x+1) − 3^(2x−1))

Answer: 3

Factor the numerator: 9^(x+1) − 9^x = 9^x(9 − 1) = 8 · 9^x = 8 · 3^(2x). Factor the denominator: 3^(2x+1) − 3^(2x−1) = 3^(2x−1)(3² − 1) = 8 · 3^(2x−1). Dividing gives 8 · 3^(2x) / (8 · 3^(2x−1)) = 3^(2x) / 3^(2x−1) = 3^1 = 3.

Rationalize and fully simplify: (√5 + √3) / (√5 − √3)

Answer: 4 + √15

Multiply numerator and denominator by the conjugate (√5 + √3): the result is (√5 + √3)² / ((√5)² − (√3)²) = (5 + 2√15 + 3) / (5 − 3) = (8 + 2√15) / 2 = 4 + √15. Option D is the unsimplified numerator; option C uses √30 instead of √15; option A has a sign error.

Solve for x: 4^x − 6 · 2^x + 8 = 0

Answer: x = 1 or x = 2

Let u = 2^x, so 4^x = (2^x)² = u². The equation becomes u² − 6u + 8 = 0, factoring as (u − 2)(u − 4) = 0. Then u = 2 gives 2^x = 2 so x = 1; u = 4 gives 2^x = 4 so x = 2. Option A confuses the u-values with x-values.

Solve for x: √(3x − 2) = x − 2. Which of the following correctly states all valid solutions?

Answer: One solution: x = 6 only

Squaring both sides: 3x − 2 = (x − 2)² = x² − 4x + 4, giving x² − 7x + 6 = 0, so (x − 1)(x − 6) = 0. Check x = 1: the right-hand side x − 2 = −1 < 0, which is impossible for a square root — x = 1 is extraneous. Check x = 6: √(18 − 2) = √16 = 4 = 6 − 2 ✓. Only x = 6 is valid.

Given that 3^a = 5, find the exact value of 9^(2a − 1).

Answer: 625/9

Write 9^(2a−1) = (3²)^(2a−1) = 3^(4a−2) = 3^(4a) ÷ 3² = (3^a)^4 ÷ 9 = 5^4 ÷ 9 = 625/9. Option B squares instead of raising to the 4th power; option C uses 4a+2 in the exponent; option D divides by 3 rather than 9.

If x = (√5 − √3) / (√5 + √3), what is the exact value of x + 1/x?

Answer: 8

x + 1/x = (√5−√3)/(√5+√3) + (√5+√3)/(√5−√3). Writing over a common denominator: [(√5−√3)² + (√5+√3)²] / [(√5+√3)(√5−√3)]. Numerator: (5 − 2√15 + 3) + (5 + 2√15 + 3) = 16. Denominator: 5 − 3 = 2. So x + 1/x = 16/2 = 8. The ±2√15 terms cancel completely.