Difference between revisions of "2023 AMC 12B Problems/Problem 7"

Latest revision as of 21:08, 19 August 2024

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: $0<a<2$ 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

Solution 3 (3 degree polynominal graph)

for $\frac{a(2-a)}{a-3} >0$ ,

transform it into a(a-2)(a-3) <0 ,

use the following graph to quickly confirm

1) a < 0 or

2) 2 < a <3

~luckuso

Video Solution

https://youtu.be/GGdJJzzbivM

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

@@ Line 1: / Line 1: @@
+==Problem==
+For how many integers <math>n</math> does the expression<cmath>\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}</cmath>represent a real number, where log denotes the base <math>10</math> logarithm?
+<math>\textbf{(A) }900 \qquad \textbf{(B) }3\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2  \qquad \textbf{(E) }901</math>
 ==Solution==
@@ Line 25: / Line 30: @@
 Putting all cases together, the total number of solutions is
-\boxed{\textbf{(E) 901}}.
+<math>\boxed{\textbf{(E) 901}}</math>.
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Solution (Solution 1 for dummies)==
+Notice <math>\log(n^2)</math> can be written as <math>2\log(n)</math>. Setting <math>a=\log(n)</math>, the equation becomes <math>\sqrt{\frac{2a-a^2}{a-3}}</math> which can be written as <math>\sqrt{\frac{a(2-a)}{a-3}}</math>
+Case 1: <math>a \ge 3</math>
+The expression is undefined when <math>a=3</math>. For <math>a>3</math>, it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions.
+Case 2: <math>2 \le a<3</math>
+For <math>a=2</math>, the numerator is zero, giving us a valid solution. When <math>a>2</math>, 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 <math>10^2</math> and <math>10^3-1</math>. There are 900 solutions here.
+Case 3: <math>0<a<2</math>
+The numerator will be positive but the denominator is negative, no real solutions exist.
+Case 4: <math>a=0</math>
+The expression evaluates to zero, <math>1</math> valid solution exists.
+Case 5: <math>a<0</math>
+All values for <math>a<0</math> requires <math>0<n<1</math>, no integer solutions exist.
+Adding up the cases:
+<math>900+1=\boxed{\textbf{(E) 901}}</math>
+~woeIsMe
+typesetting: paras
+==Solution 3 (3 degree polynominal graph) ==
+for <math>\frac{a(2-a)}{a-3} >0  </math> ,
+transform it into a(a-2)(a-3) <0 ,
+use the following graph to quickly confirm
+) a < 0 or
+) 2 < a <3
+~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+[[Image:2023_AMC12B_P7.PNG|thumb|center|600px]]
+==Video Solution==
+https://youtu.be/GGdJJzzbivM
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==See Also==
+{{AMC12 box|year=2023|ab=B|num-b=6|num-a=8}}
+{{MAA Notice}}

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search