2023 AMC 12B Problems/Problem 7

Problem

For how many integers $n$ does the expression $\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}$ represent a real number, where log denotes the base $10$ logarithm?

$\textbf{(A) }900 \qquad \textbf{(B) }2\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2 \qquad \textbf{(E) }901$

Solution

We have $\begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*}$

Because $n$ is an integer and $\log n$ is well defined, $n$ must be a positive integer.

Case 1: $n = 1$ or $10^2$ .

The above expression is 0. So these are valid solutions.

Case 2: $n \neq 1, 10^2$ .

Thus, $\log n > 0$ and $2 - \log n \neq 0$ . To make the above expression real, we must have $2 < \log n < 3$ . Thus, $100 < n < 1000$ . Thus, $101 \leq n \leq 999$ . Hence, the number of solutions in this case is 899.

Putting all cases together, the total number of solutions is \boxed{\textbf{(E) 901}}.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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2023 AMC 12B Problems/Problem 7

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