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

m (→‎Solution)
Line 34: Line 34:
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 +
==Solution (Solution 1 for dummies)==
 +
 +
Notice log(n^2) can be written as 2log(n). Setting a=log(n), the equation becomes sqrt((2a-a^2)/(a-3)) which can be written as sqrt((a(a-2))/(a-3))
 +
 +
Case 1: a>=3
 +
The expression is undefined when a=3. For a>3, it is trivial to see that the denominator is negative and the numerator is positive, thus resulting in no real solutions.
 +
 +
Case 2: 2<=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 negative but the denominator is positive, 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=<math>\boxed{\textbf{(E) 901}}</math>
 +
 +
Side Note: Someone please fix this to make it readable. I don't know anything about latex.
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2023|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:52, 15 November 2023

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 2log(n). Setting a=log(n), the equation becomes sqrt((2a-a^2)/(a-3)) which can be written as sqrt((a(a-2))/(a-3))

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

Case 2: 2<=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 negative but the denominator is positive, 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=$\boxed{\textbf{(E) 901}}$

Side Note: Someone please fix this to make it readable. I don't know anything about latex.

See Also

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

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