Difference between revisions of "2020 AIME II Problems/Problem 3"
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+ | https://www.youtube.com/watch?v=2TSNY2DDUbQ&t=3s ~ MathEx | ||
==See Also== | ==See Also== |
Revision as of 19:45, 19 June 2020
Contents
Problem
The value of that satisfies can be written as , where and are relatively prime positive integers. Find .
Solution
Let . Based on the equation, we get and . Expanding the second equation, we get . Substituting the first equation in, we get , so . Taking the 100th root, we get . Therefore, , so and the answer is . ~rayfish
Easiest Solution
Recall the identity (which is easily proven using exponents or change of base) Then this problem turns into Divide from both sides. And we are left with .Solving this simple equation we get ~mlgjeffdoge21
Solution 2
Because we have that or Since and thus resulting in or We remove the base 3 logarithm and the power of 2 to yield or
Our answer is ~ OreoChocolate
Solution 3 (Official MAA)
Using the Change of Base Formula to convert the logarithms in the given equation to base yields Canceling the logarithm factors then yieldswhich has solution The requested sum is .
Video Solution
https://youtu.be/lPr4fYEoXi0 ~ CNCM
Video Solution 2
https://www.youtube.com/watch?v=x0QznvXcwHY?t=528
~IceMatrix
Video Solution 3
~avn
Video Solution 4
https://www.youtube.com/watch?v=2TSNY2DDUbQ&t=3s ~ MathEx
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.