# 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?

$\textbf{(A) }900 \qquad \textbf{(B) }3\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2 \qquad \textbf{(E) }901$

## 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)

## Solution (Solution 1 for dummies)

Notice $\log(n^2)$ can be written as $2\log(n)$. Setting $a=\log(n)$, the equation becomes $\sqrt{\frac{2a-a^2}{a-3}}$ which can be written as $\sqrt{\frac{a(2-a)}{a-3}}$

Case 1: $a \ge 3$ The expression is undefined when $a=3$. For $a>3$, it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions.

Case 2: $2 \le a<3$ For $a=2$, the numerator is zero, giving us a valid solution. When $a>2$, both the denominator and numerator are negative so all real values of a in this interval is a solution to the equation. All integers of n that makes this true are between $10^2$ and $10^3-1$. There are 900 solutions here.

Case 3: $0 The numerator will be positive but the denominator is negative, no real solutions exist.

Case 4: $a=0$ The expression evaluates to zero, $1$ valid solution exists.

Case 5: $a<0$ All values for $a<0$ requires $0, no integer solutions exist.

Adding up the cases: $900+1=\boxed{\textbf{(E) 901}}$

~woeIsMe typesetting: paras

## Video Solution

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