AMC 10 · 2023 · #19

Grade 11 algebra

Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery

logarithm-propertiesquadratic-equationsvieta-formulasdomain-restriction convert-to-algebraidentify-subproblems ↑ Prerequisites: logarithm-propertiesquadratic-equations

📏 Long solution 💡 3 insights

Problem

Three logarithms appear, all with the same argument but different bases. The equation says the first two multiply to give the third. Find every value making this true, then multiply those values together.

Pick an answer.

(A)

$(\log_{2023}7\cdot \log_{2023}289)^2$

(B)

$\log_{2023}7\cdot \log_{2023}289$

(C)

(D)

$\log_{7}2023\cdot \log_{289}2023$

(E)

$(\log_7 2023\cdot\log_{289} 2023)^2$

How to solve

Strategy Introduce a Variable

The unknown x is buried inside three different bases, which is the worst place for it to sit, because base arithmetic has no useful rules. Tool #15 fixes that first: flipping each logarithm upside down, using log_bN = 1/log_Nb, moves every x out of a base and into an argument, and lands all three logarithms on the single common base 2023. Now the log rules apply. Tool #4 then does the real work. Setting u = log₂₀₂₃x turns log₂₀₂₃(7x) into log₂₀₂₃₇ + u and its two siblings into the same shape, so the whole equation becomes a quadratic in u. The hidden gift is that 2023 = 7 · 289, which makes the two constants log₂₀₂₃₇ and log₂₀₂₃₂₈₉ add to exactly 1 — and that single fact is what makes the linear terms cancel. The last move is the one that decides the problem. The question wants a product of x values, but the quadratic lives in u. Since x = 2023^u, multiplying the x values means adding the u values, so Tool #16 says to look at the SUM of the roots by Vieta, never the product. Tool #14 confirms both roots are real before Vieta is trusted, and Tool #3 confirms neither root is one of the forbidden values.

1STEP 1

List what is allowed

A base of one is not allowed.

x ＞ 0, x ≠ 1/7, x ≠ 1/289, x ≠ 1/2023

2STEP 2

Flip onto one base

Flip every logarithm onto one base.

1/(log₂₀₂₃(7x)·log₂₀₂₃(289x)) = 1/log₂₀₂₃(2023x) ⟹ log₂₀₂₃(7x)·log₂₀₂₃(289x) = log₂₀₂₃(2023x)

3STEP 3

Spot the base product

The two bases multiply to the third base.

u=log₂₀₂₃x, a=log₂₀₂₃₇, b=log₂₀₂₃₂₈₉ ⟹ a+b=log₂₀₂₃(7 · 289)=1, (a+u)(b+u)=1+u

4STEP 4

Expand and cancel

Expanding cancels the linear terms.

ab+(a+b)u+u² = 1+u ⟹ ab+u+u² = 1+u ⟹ u² = 1-ab

5STEP 5

Confirm two real roots

Two real roots exist.

ab ≤ ((a+b)/2)²=1/4 ⟹ 1-ab ≥ 3/4 ＞ 0; u₁=√(1-ab), u₂=-√(1-ab), u₁+u₂=0

6STEP 6

The product is a sum of exponents

The roots sum to zero, so the product is one.

x₁x₂ = 2023^u₁ · 2023^u₂ = 2023^ u₁+u₂ = 2023⁰ = 1

7STEP 7

Check legality and answer

Both are legal, so the answer is 1.

u=-a → a²=1-ab → a(a+b)=1 → a=1, false; likewise u=-b and u=-1 fail ⟹ x₁x₂ = 1 → (C)

Answer

Solve numerically and check by hand. With a=log₂₀₂₃₇ ≈ 0.25563 and b=log₂₀₂₃₂₈₉ ≈ 0.74437, indeed a+b=1 and ab ≈ 0.19028, so √(1-ab) ≈ 0.89984. The two solutions are x₁ = 2023⁰.89984 ≈ 943.7993 and x₂ = 2023⁻⁰.89984 ≈ 0.00105955. Substituting x₁ into the original equation gives left side ≈ 0.5263590839 and right side ≈ 0.5263590839; substituting x₂ gives ≈ 9.9843963841 on both sides. Both are positive and neither equals 1/7≈ 0.142857, 1/289≈ 0.003460, or 1/2023≈ 0.000494, so both are admissible. Their product is 943.7993 × 0.00105955 = 1.000000. Note also that x₂ is exactly 1/x₁: the solution set is a reciprocal pair, which is the concrete face of u₁+u₂=0. Now the decoys. Writing a b for short, choice (B) is ab ≈ 0.1903, choice (A) is (ab)² ≈ 0.0362, choice (D) is log₇₂₀₂₃·log₂₈₉₂₀₂₃ = 1/ab ≈ 5.2554, and choice (E) is 1/(ab)² ≈ 27.6190. All four are the leftover quantity ab dressed up four ways — and ab is genuinely the thing that appears in u² = 1-ab, which is why it is such a tempting place to stop. The point of the problem is that ab shifts the two roots apart or together but never breaks their u₁+u₂=0 symmetry, so it cannot survive into the product. Only choice (C) reflects that cancellation.

💡Key takeaway

When you solve for the logarithm of x instead of x itself, multiplying the answers together turns into adding the roots — so use Vieta's sum, not Vieta's product.

List what x is allowed to be
Flip every log onto base 2023
Name the log of x, and use 2023 = 7 times 289
Expand and watch the u terms cancel
Confirm two real roots exist
Product of x means sum of u
Check both solutions are legal, then answer