Difference between revisions of "2010 AIME II Problems/Problem 5"
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<math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math> | <math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math> | ||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=Ix6czB_A_Js&t=2s | ||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=4|num-a=6|n=II}} | {{AIME box|year=2010|num-b=4|num-a=6|n=II}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:55, 21 July 2020
Problem
Positive numbers , , and satisfy and . Find .
Solution
Using the properties of logarithms, by taking the log base 10 of both sides, and by using the fact that .
Through further simplification, we find that . It can be seen that there is enough information to use the formula , as we have both and , and we want to find .
After plugging in the values into the equation, we find that is equal to .
However, we want to find , so we take the square root of , or .
Solution 2
Video solution
https://www.youtube.com/watch?v=Ix6czB_A_Js&t=2s
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.