AMC 10 · 2003 · #17

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
logarithm-propertiessystems-of-equationslinear-equations-two-var convert-to-algebraidentify-subproblems ↑ Prerequisites: logarithm-properties
📏 Medium solution 💡 2 insights
Problem
Two positive numbers x and y satisfy log(xy³)=1 and log(x²y)=1. Using only those two facts, find the value of log(xy).

Pick an answer.

(A)
$-\frac 12$
(B)
0
(C)
$\frac 12$
(D)
$\frac 35$
(E)
1
How to solve
Strategy Introduce a Variable

The logs look complicated only because x and y sit inside them. Break each given apart with the product and power rules and every equation becomes linear in the two quantities log x and log y (Tool #13). Rename those two quantities a and b so the problem stops looking like logarithms and starts looking like a two-by-two linear system (Tool #4, Tool #15). The target log(xy) is then a+b — a combination, not an individual unknown — so instead of solving for a and b separately, hunt for the multiples of the two equations that add up to a multiple of a+b (Tool #16). Solving the system fully afterwards is a cheap independent check.

1STEP 1

Break the logs into pieces

Splitting the products turns both givens into plain sums of two logarithms.

log(xy³)=log x+3log y, log(x²y)=2log x+log y, log(xy)=log x+log y
2STEP 2

Rename to a linear system

Renaming them makes an ordinary linear system, with no logarithms in sight.

a=log x, b=log y: a+3b=1, 2a+b=1, find a+b
3STEP 3

Aim straight at the sum

Doubling one and adding the other makes both coefficients five, giving 5(a+b) = 3.

2(2a+b)+(a+3b)=4a+2b+a+3b=5a+5b=2+1=3
4STEP 4

Divide, then confirm the pair exists

Dividing gives 3/5, and solving outright confirms such x and y exist, choice (D).

a+b=3/5; a=2/5, b=1/5 → x=10²/5, y=10¹/5
Answer
3/5
Substitute the recovered values back into the original statements: xy³=10²/5 · 10³/5=10¹, so log(xy³)=1; and x²y=10⁴/5 · 10¹/5=10¹, so log(x²y)=1. Both givens hold exactly, and xy=10²/5+1/5=10³/5 gives log(xy)=3/5. A size check agrees: xy sits between 10¹/2 and 10¹, so its log must be between 1/2 and 1 — which rules out (A), (B), and (C) at a glance and leaves only (D) and (E). The trap is (E) 1: both givens equal 1, so it is tempting to assume the symmetric-looking log(xy) is 1 too, but xy³ and x²y each carry four factors' worth of x and y combined while xy carries only two, so log(xy) must be strictly smaller.
💡Key takeaway

Logs turn products into sums, so log x and log y become two ordinary unknowns — and when only their sum is wanted, combine the equations to build that sum instead of solving for each one.

  • Break the logs into pieces
  • Rename to a linear system
  • Aim straight at the sum
  • Divide, then confirm the pair exists

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