AMC 10 · 2016 · #16

Grade 11 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

logarithm-propertiesdomain-restrictionquadratic-equationscombinations-basic systematic-enumerationconvert-to-algebra ↑ Prerequisites: logarithm-propertiesquadratic-equations

📏 Medium solution 💡 3 insights

Problem

Four logarithmic curves are drawn together on one set of axes. Count the points on at least two of them.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Four logarithms with four different base-and-input arrangements look like four unrelated curves, but change of base shows they are all built from one number. Tool #4 (Introduce a Variable) names that number t=log₃ x, and the four curves collapse to t, 1/t, -t, and -1/t — tool #9 (Solve an Easier Related Problem) in action, since two lines and two hyperbolas are far easier to intersect than four logarithms. Tool #2 (Make a Systematic List) then walks every one of the C(4, 2)=6 pairs so no crossing is missed and none is double-counted, and tool #3 (Eliminate Possibilities) discards the pairs whose equation has no real solution. Finally tool #15 (Organize Information in More Ways) regroups the surviving solutions by x-value, which is what makes it obvious that the five points are genuinely different.

1STEP 1

List the curves and their domains

Each curve carries its own allowed inputs.

y=log₃ x, y=log₁/3 x: x ＞ 0; y=log_x 3, y=log_x 1/3: x ＞ 0, x ≠ 1

2STEP 2

Rename log₃ x as t

One rename makes all four simple.

t=log₃ x: y=t, y=1/t, y=-t, y=-1/t

3STEP 3

Check the rename loses nothing

The rename loses nothing.

x ＞ 0⇔ tinR, x=3^t; x ≠ 1⇔ t ≠ 0

4STEP 4

Write all six pair equations

There are six pairs to check.

t=1/t, t=-t, t=-1/t, 1/t=-t, 1/t=-1/t, -t=-1/t

5STEP 5

Solve them and discard the impossible

Some pairs have no real solution.

t²=1→ t=± 1; 2t=0→ t=0; t²=-1: no real solution; 2/t=0: no solution

6STEP 6

Convert back and count

Converting back gives 5 points, choice (D).

(3,1), (3,-1), (1/3,-1), (1/3,1), (1,0) → 5 → (D)

Answer

Direct substitution confirms the five points without any rename. At x=3 the four heights are log₃ 3=1, log₃ 3=1, log₁/3 3=-1, log₃ 1/3=-1, giving the two points (3,1) and (3,-1). At x=1/3 they are -1, -1, 1, 1, giving (1/3,-1) and (1/3,1). At x=1 curves 1 and 3 are both 0 while curves 2 and 4 are undefined, giving (1,0). That is 5 points, matching choice (D). Nothing was missed: for any x ＞ 0 with x ≠ 1, writing u=log₃ x ≠ 0, the four heights are the numbers u, 1/u, -u, -1/u, and two of these can be equal only when u=1/u, that is u=± 1, that is x=3 or x=1/3; the other possible coincidences need u=-u or u²=-1, both impossible for a nonzero real u. So 5 is exactly right, and choice (E) 6 would require a crossing that provably does not exist.

💡Key takeaway

Give log₃ x the short name t: the four graphs turn into t, 1/t, -t, and -1/t, and checking all six pairs shows they cross at exactly 5 points.

List the curves and their domains
Rename log₃ x as t
Check the rename loses nothing
Write all six pair equations
Solve them and discard the impossible
Convert back and count