AMC 10 · 2019 · #12

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta
logarithm-propertiesexponentssystems-of-equationssymmetric-polynomials convert-to-algebrasymmetric-polynomialsidentify-subproblems ↑ Prerequisites: logarithm-propertiesexponents
📏 Medium solution 💡 3 insights
Problem
Two positive numbers, neither equal to 1, satisfy two conditions: the base-two logarithm of the first equals the logarithm of sixteen in the base of the second, and their product is 64. Find the square of the base-two logarithm of their quotient.

Pick an answer.

(A)
$\frac{25}{2}$
(B)
20
(C)
$\frac{45}{2}$
(D)
25
(E)
32
How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable): every quantity in this problem is a base-2 logarithm in disguise, so naming a = log₂{x} and b = log₂{y} turns three log statements into three plain algebra statements. Tool #13 (Convert to Algebra): change-of-base and the log rules convert both givens and the target into equations in a and b. Tool #15 (Reorganize): the target (a-b)² never needs a and b separately — rewriting it as (a+b)² - 4ab lets the two givens plug straight in, so no quadratic ever has to be solved.

1STEP 1

Name the two logarithms

Give the two logarithms names.

a = log₂{x}, b = log₂{y}, x = 2^a, y = 2^b, a ≠ 0, b ≠ 0
2STEP 2

Turn the first given into ab = 4

The first condition fixes their product.

log_y{16} = log₂{16}/log₂{y} = 4/b ⟹ a = 4/b ⟹ ab = 4
3STEP 3

Turn the second given into a + b = 6

The second fixes their sum.

xy = 2^a · 2^b = 2^a+b = 64 = 2⁶ ⟹ a + b = 6
4STEP 4

Rewrite the target as (a - b)²

The target is the squared difference.

(log₂{x/y})² = (a - b)², with a+b = 6 and ab = 4
5STEP 5

Use sum and product, skip solving

Without solving, sum and product give 20.

(a-b)² = (a+b)² - 4ab = 6² - 4(4) = 36 - 16 = 20 → (B) 20
Answer
20
Solve for a and b explicitly as a check. With a + b = 6 and ab = 4, the numbers a and b are the roots of t² - 6t + 4 = 0, so t = 3 ± √(5). Then a - b = ± 2√(5) and (a-b)² = 4 · 5 = 20, matching. The values are legal: 3 ± √(5) are both nonzero, so x = 2³+√(5) and y = 2³-√(5) are positive and neither equals 1; their product is 2⁶ = 64 as required. A size check also helps: a and b are near 5.24 and 0.76, so a - b ≈ 4.47 and its square is about 20 — nowhere near 32, and not a half-integer like 25/2 or 45/2.
💡Key takeaway

This AMC 12 problem needs only Grade 11 logarithm rules: name a = log₂{x} and b = log₂{y}, and the two givens become ab = 4 and a + b = 6. Since (a-b)² = (a+b)² - 4ab = 36 - 16, you never have to find a and b themselves. The answer is (B).

  • Name the two logarithms
  • Turn the first given into ab = 4
  • Turn the second given into a + b = 6
  • Rewrite the target as (a - b)²
  • Use sum and product, skip solving

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