2023 AMC 12B Problems/Problem 7
Contents
Problem
For how many integers does the expressionrepresent a real number, where log denotes the base logarithm?
Solution
We have
Because is an integer and is well defined, must be a positive integer.
Case 1: or .
The above expression is 0. So these are valid solutions.
Case 2: .
Thus, and . To make the above expression real, we must have . Thus, . Thus, . Hence, the number of solutions in this case is 899.
Putting all cases together, the total number of solutions is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution (Solution 1 for dummies)
Notice can be written as . Setting , the equation becomes which can be written as
Case 1: The expression is undefined when . For , it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions.
Case 2: For , the numerator is zero, giving us a valid solution. When , 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 and . There are 900 solutions here.
Case 3: The numerator will be positive but the denominator is negative, no real solutions exist.
Case 4: The expression evaluates to zero, valid solution exists.
Case 5: All values for requires , no integer solutions exist.
Adding up the cases:
~woeIsMe typesetting: paras
Solution 3 (3 degree polynominal graph)
for ,
transform it into a(a-2)(a-3) <0 ,
use the following graph to quickly confirm
1) a < 0 or
2) 2 < a <3
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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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