2023 AMC 12B Problem 7

Attempt Problem 7 of the 2023 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

For how many integers nn does the expression
log⁡(n2)−(log⁡n)2log⁡n−3 \sqrt{\frac{\log(n^2)-(\log n)^2}{\log n-3}}
represent a real number, where log⁡\log denotes the base 1010 logarithm?
900900
33
902902
22
901901
Answer: E
Concepts:logarithminequality
Small hint:
The expression is real exactly when the fraction under the square root is ≥0\ge 0
Big hint:
Let L=log⁡n;L=\log n; the fraction is L(2−L)L−3≥0,\dfrac{L(2-L)}{L-3}\ge 0, and nn must be a positive integer
Solution:
Write L=log⁡n.L=\log n. Then log⁡(n2)−(log⁡n)2\log(n^2)-(\log n)^2 =2L−L2=2L-L^2 =L(2−L),=L(2-L), and the fraction is L(2−L)L−3.\dfrac{L(2-L)}{L-3}. A sign chart shows this is ≥0\ge 0 exactly when L≤0L\le 0 or 2≤L<3.2\le L\lt 3. Since nn is a positive integer, L≤0L\le 0 forces n=1,n=1, while 2≤L<32\le L\lt 3 gives 100≤n≤999,100\le n\le 999, which is 900900 values. In total 1+900=901.1+900=901.
Thus, the correct answer is E.
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