2022 AIME II Problems/Problem 4
Contents
Problem
There is a positive real number not equal to either or such thatThe value can be written as , where and are relatively prime positive integers. Find .
Solution 1
Define to be , what we are looking for. Then, by the definition of the logarithm, Dividing the first equation by the second equation gives us , so by the definition of logs, . This is what the problem asked for, so the fraction gives us .
~ihatemath123
Solution 2
We could assume a variable which equals to both and .
So that and
Express as:
Substitute to :
Thus, , where and .
Therefore, .
Solution 3
We have
We have
Because , we get
We denote this common value as .
By solving the equality , we get .
By solving the equality , we get .
By equating these two equations, we get
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4 (Solution 1 with more reasoning)
Let be the exponent such that and . Dividing, we get Thus, we see that , so the answer is .
~A_MatheMagician
Solution 5
By the change of base rule, we have , or . We also know that if , then this also equals . We use this identity and find that . The requested sum is
~MathIsFun286
Solution 6
By change of base formula, Thus, The requested answer is .
~ adam_zheng
Video Solution
https://www.youtube.com/watch?v=4qJyvyZN630
Video Solution by Power of Logic
~Hayabusa1
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.