Answer :
To determine for how many integers [tex]\( n \)[/tex] the expression
[tex]\[ \sqrt{\frac{\log \left(n^2\right) - (\log n)^2}{\log n - 3}} \][/tex]
is a real number, let's analyze the expression step by step.
1. Start by simplifying the expression inside the square root:
[tex]\[ \sqrt{\frac{\log \left(n^2\right) - (\log n)^2}{\log n - 3}} \][/tex]
Recall that [tex]\( \log(n^2) = 2\log(n) \)[/tex]. Substituting this into the expression, we get:
[tex]\[ \sqrt{\frac{2\log(n) - (\log n)^2}{\log n - 3}} \][/tex]
2. Let's introduce a new variable to simplify the notation. Let [tex]\( y = \log n \)[/tex]. Then the expression becomes:
[tex]\[ \sqrt{\frac{2y - y^2}{y - 3}} \][/tex]
3. The expression inside the square root, [tex]\(\frac{2y - y^2}{y - 3}\)[/tex], must be non-negative for the entire expression to be a real number. So, we need to solve the inequality:
[tex]\[ \frac{2y - y^2}{y - 3} \geq 0 \][/tex]
4. Factor the numerator to get:
[tex]\[ 2y - y^2 = y(2 - y) \][/tex]
Hence, the inequality can be rewritten as:
[tex]\[ \frac{y(2 - y)}{y - 3} \geq 0 \][/tex]
5. Determine the critical points where the expression could potentially change sign. These are the points where the numerator or denominator is zero:
- [tex]\( y = 0 \)[/tex] (numerator)
- [tex]\( y = 2 \)[/tex] (numerator)
- [tex]\( y = 3 \)[/tex] (denominator)
6. Analyze the sign changes around these critical points:
- For [tex]\( y < 0 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is positive, and [tex]\( y - 3 \)[/tex] is negative, so the overall fraction is negative.
- For [tex]\( 0 < y < 2 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is positive, and [tex]\( y - 3 \)[/tex] is negative, so the overall fraction is negative.
- For [tex]\( 2 < y < 3 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is negative, and [tex]\( y - 3 \)[/tex] is negative, so the overall fraction is positive.
- For [tex]\( y > 3 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is negative, and [tex]\( y - 3 \)[/tex] is positive, so the overall fraction is negative.
7. Therefore, the expression [tex]\(\frac{y(2 - y)}{y - 3}\)[/tex] is only non-negative when [tex]\( 2 < y < 3 \)[/tex].
8. Since [tex]\( y = \log n \)[/tex], we need:
[tex]\[ 2 < \log n < 3 \][/tex]
9. Converting this inequality back in terms of [tex]\( n \)[/tex]:
[tex]\[ 10^2 < n < 10^3 \][/tex]
Thus:
[tex]\[ 100 < n < 1000 \][/tex]
10. Thus, [tex]\( n \)[/tex] must be an integer between 101 and 999. The total number of integers [tex]\( n \)[/tex] that satisfy this condition is:
[tex]\[ 999 - 101 + 1 = 899 \][/tex]
Therefore, the expression represents a real number for 899 integers [tex]\( n \)[/tex].
[tex]\[ \sqrt{\frac{\log \left(n^2\right) - (\log n)^2}{\log n - 3}} \][/tex]
is a real number, let's analyze the expression step by step.
1. Start by simplifying the expression inside the square root:
[tex]\[ \sqrt{\frac{\log \left(n^2\right) - (\log n)^2}{\log n - 3}} \][/tex]
Recall that [tex]\( \log(n^2) = 2\log(n) \)[/tex]. Substituting this into the expression, we get:
[tex]\[ \sqrt{\frac{2\log(n) - (\log n)^2}{\log n - 3}} \][/tex]
2. Let's introduce a new variable to simplify the notation. Let [tex]\( y = \log n \)[/tex]. Then the expression becomes:
[tex]\[ \sqrt{\frac{2y - y^2}{y - 3}} \][/tex]
3. The expression inside the square root, [tex]\(\frac{2y - y^2}{y - 3}\)[/tex], must be non-negative for the entire expression to be a real number. So, we need to solve the inequality:
[tex]\[ \frac{2y - y^2}{y - 3} \geq 0 \][/tex]
4. Factor the numerator to get:
[tex]\[ 2y - y^2 = y(2 - y) \][/tex]
Hence, the inequality can be rewritten as:
[tex]\[ \frac{y(2 - y)}{y - 3} \geq 0 \][/tex]
5. Determine the critical points where the expression could potentially change sign. These are the points where the numerator or denominator is zero:
- [tex]\( y = 0 \)[/tex] (numerator)
- [tex]\( y = 2 \)[/tex] (numerator)
- [tex]\( y = 3 \)[/tex] (denominator)
6. Analyze the sign changes around these critical points:
- For [tex]\( y < 0 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is positive, and [tex]\( y - 3 \)[/tex] is negative, so the overall fraction is negative.
- For [tex]\( 0 < y < 2 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is positive, and [tex]\( y - 3 \)[/tex] is negative, so the overall fraction is negative.
- For [tex]\( 2 < y < 3 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is negative, and [tex]\( y - 3 \)[/tex] is negative, so the overall fraction is positive.
- For [tex]\( y > 3 \)[/tex]: [tex]\( y(2 - y) \)[/tex] is negative, and [tex]\( y - 3 \)[/tex] is positive, so the overall fraction is negative.
7. Therefore, the expression [tex]\(\frac{y(2 - y)}{y - 3}\)[/tex] is only non-negative when [tex]\( 2 < y < 3 \)[/tex].
8. Since [tex]\( y = \log n \)[/tex], we need:
[tex]\[ 2 < \log n < 3 \][/tex]
9. Converting this inequality back in terms of [tex]\( n \)[/tex]:
[tex]\[ 10^2 < n < 10^3 \][/tex]
Thus:
[tex]\[ 100 < n < 1000 \][/tex]
10. Thus, [tex]\( n \)[/tex] must be an integer between 101 and 999. The total number of integers [tex]\( n \)[/tex] that satisfy this condition is:
[tex]\[ 999 - 101 + 1 = 899 \][/tex]
Therefore, the expression represents a real number for 899 integers [tex]\( n \)[/tex].
