AMC 10 · 2023 · #7

Grade 11 algebra
logarithm-propertiesdomain-restrictionsign-analysisconsecutive-integer-count identify-subproblemsconvert-to-algebrasign-analysis ↑ Prerequisites: logarithm-propertiessign-analysis
📏 Medium solution 💡 3 insights
Problem
Base-ten logarithms are used throughout. Count how many integers make a certain square-root expression a real number.

Pick an answer.

(A)
900
(B)
3
(C)
902
(D)
2
(E)
901
How to solve
Strategy Introduce a Variable

Nothing here has to be computed — the whole problem is deciding when the expression is even allowed to exist. Tool #7 (Identify Subproblems) splits that into three independent demands: the logarithms must be defined, the denominator must not vanish, and the radicand must not be negative. Tool #4 (Introduce a Variable) does the heavy lifting: the symbol log n shows up three times, so naming it t turns a logarithm question into the rational inequality (t(2-t))/(t-3) ≥ 0, which is ordinary algebra. Tool #2 (Make a Systematic List) handles that inequality with a sign chart — the roots cut the line into four intervals and one test value settles each. Tool #14 (Extreme Principle) is what separates the right answer from the near misses: every wrong choice in this problem comes from mishandling a boundary point, so t = 0, t = 2, and t = 3 each get checked by hand.

1STEP 1

List every condition for realness

List every condition for realness.

n > 0, log n ≠ 3, (log (n²) - (log n)²)/(log n - 3) ≥ 0
2STEP 2

Name the logarithm t

Give the logarithm a name.

t = log n ⟹ (2t - t²)/(t - 3) = (t(2-t))/(t-3) ≥ 0
3STEP 3

Build a sign chart

Build a sign chart.

t=-1: (-1)(3)/-4 = 3/4 > 0; t=1: (1)(1)/-2 = -1/2 < 0; t=2.5: (2.5)(-0.5)/-0.5 = 2.5 > 0; t=4: (4)(-2)/1 = -8 < 0
4STEP 4

Decide the three border points

Remove where the denominator vanishes.

t ≤ 0 or 2 ≤ t < 3
5STEP 5

Translate t back to n

Translate back to integers.

t ≤ 0 → 0 < n ≤ 1; 2 ≤ t < 3 → 100 ≤ n < 1000
6STEP 6

Count the integers

Counting gives 901.

1 + (999 - 100 + 1) = 1 + 900 = 901 → (E)
Answer
901
Spot-check the four values that decide the count. For n = 1: log 1 = 0, so the radicand is (0 - 0)/(0 - 3) = 0 and the expression is √(0) = 0, real — so n = 1 counts. For n = 100: log 100 = 2, radicand (4 - 4)/(2 - 3) = 0, again real. For n = 99: log 99 ≈ 1.9956, radicand ≈ (3.9913 - 3.9825)/-1.0044 ≈ -0.0088, negative, so 99 is correctly excluded. For n = 1000: the denominator is 0, so nothing is defined. For n = 1001: log n slightly exceeds 3, making the numerator negative and the denominator positive, so the radicand is negative. The three wrong-looking choices are exactly the boundary mistakes: 900 drops n = 1, 902 keeps both n = 1000 and one more forbidden value, and the tiny answers 2 and 3 come from only counting the perfect powers of 10. Answer 901 survives all of these checks.
💡Key takeaway

Give the repeated log n a single name, then remember a square root only needs its inside to be ≥ 0 — so the spots where the top is zero stay in, and the spot where the bottom is zero drops out.

  • List every condition for realness
  • Name the logarithm t
  • Build a sign chart
  • Decide the three border points
  • Translate t back to n
  • Count the integers