AMC 10 · 2007 · #23

Grade 11 algebrageometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

logarithm-propertiesexponentscoordinate-geometryquadratic-equations convert-to-algebrawork-backwards ↑ Prerequisites: logarithm-properties

📏 Long solution 💡 4 insights

Problem

A square of area 36 has three of its corners on three logarithm curves with the same base. Find that base.

Pick an answer.

(A)

$\sqrt [6]{3}$

(B)

$\sqrt {3}$

(C)

$\sqrt [3]{6}$

(D)

$\sqrt {6}$

(E)

How to solve

Strategy Organize Information in More Ways

The three graphs look like three separate equations, but they are one curve stretched vertically by 1, 2, and 3. Tool #15 (Organize Information in More Ways) reads that single fact in two directions. Read up a vertical line: above one x-value the three heights are L, 2L, 3L, so the vertical side BC measures |L| with no algebra at all. Read across a horizontal line: at one height the x-coordinates on the first two curves are q² and q, so the horizontal side AB turns into a quadratic. Tool #1 (Draw a Diagram) first pins the square to the axes so that 'horizontal' and 'vertical' become equations, and later builds the finished square to prove it truly exists. Tool #3 (Eliminate Possibilities) discards the root that leaves the domain of the logarithm and the base the answer list excludes. Tool #11 (Work Backwards) converts the surviving logarithm statement back into a power of a.

1STEP 1

Place the square in coordinates

Placing the square gives the corners shared coordinates.

A=(p,h), B=(q,h), C=(q,k), |p-q|=6, |k-h|=6, p ＞ 0, q ＞ 0

2STEP 2

Compare heights above the same x

Comparing heights above the same position gives one logarithm value.

L = log_aq, h = 2L, k = 3L ⟹ |k-h| = |L| = 6

3STEP 3

Compare widths at the same height

Comparing widths shows one coordinate is the square of the other.

p = a^h = a²L = (a^L)² = q² ⟹ |q² - q| = 6

4STEP 4

Solve for the shared x-coordinate

Solving picks the root inside the domain.

q²-q-6=(q-3)(q+2)=0 → q=3 (q=-2 is outside the domain); q²-q+6=0 has Δ = -23 ＜ 0. q=3, p=9

5STEP 5

Undo the logarithm

Undoing the logarithm leaves two candidate bases.

log_a3 = ± 6 ⟺ a⁶ = 3 or a⁶ = 1/3

6STEP 6

Rewrite the base as a radical

Writing them as radicals shows which is listed.

a⁶=3 → a = 3¹/6 = ⁶√(3); a⁶=1/3 → a = 3⁻¹/6 ≈ 0.83 ＜ 1

7STEP 7

Build the square and confirm

Building the square confirms the sixth root of 3, choice (A).

A=(9,12), B=(3,12), C=(3,18), D=(9,18): AB = BC = 6, area = 36 ⟹ a = ⁶√(3) ⟹ (A)

Answer

⁶√(3)

The final step draws the square explicitly, so the answer is not merely forced by necessity — the configuration is shown to exist. A size check agrees too: ⁶√(3) ≈ 1.2 is only just above 1, so log_a climbs steeply, and heights of 12 and 18 above x-values as small as 3 and 9 are exactly what such a steep logarithm produces. The other choices die on the spot, because the argument pinned a⁶ = q = 3: choice (B) gives a⁶ = 27, (C) gives a⁶ = 36, (D) gives a⁶ = 216, and (E) gives a⁶ = 46656. Only (A) lands on 3.

💡Key takeaway

The three graphs are one curve at 1, 2, and 3 times the height, so the square's vertical side measures |log_aq| while its horizontal side compares q with q².

Place the square in coordinates
Compare heights above the same x
Compare widths at the same height
Solve for the shared x-coordinate
Undo the logarithm
Rewrite the base as a radical
Build the square and confirm