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)

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: $0<a<2$ 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 $0<n<1$ , no integer solutions exist.

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

~woeIsMe typesetting: paras

Video Solution

https://youtu.be/GGdJJzzbivM

~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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2023 AMC 12B Problems/Problem 7

Contents

Problem

Solution

Solution (Solution 1 for dummies)

Video Solution

See Also