2019 AMC 12A Problems/Problem 12
Contents
Problem
Positive real numbers and satisfy and . What is ?
Solution 1
Let , so that and . Then we have .
We therefore have , and deduce . The solutions to this are .
To solve the problem, we now find ~Edits by BakedPotato66
Solution 2 (slightly simpler)
After obtaining , notice that the required answer is , as before.
Solution 3
From the given data, , or
We know that , so .
Thus , so , so .
Solving for , we obtain .
Easy resubstitution further gives . Simplifying, we obtain .
Looking back at the original problem, we have What is ?
Deconstructing this expression using log rules, we get .
Plugging in our known values, we get or .
Our answer is .
Solution 4
Multiplying the first equation by , we obtain .
From the second equation we have .
Then, .
Solution 5
Let and .
Writing the first given as and the second as , we get and .
Solving for we get .
Our goal is to find . From the above, it is equal to .
Video Solution 1
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/RdIIEhsbZKw?t=1821
~ pi_is_3.14
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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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