Difference between revisions of "2010 AIME II Problems/Problem 5"
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− | == Problem | + | == Problem == |
Positive numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy <math>xyz = 10^{81}</math> and <math>(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468</math>. Find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>. | Positive numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy <math>xyz = 10^{81}</math> and <math>(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468</math>. Find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>. | ||
== Solution == | == Solution == | ||
− | Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10} | + | Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>. |
+ | |||
+ | Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>. | ||
+ | |||
+ | After plugging in the values into the equation, we find that <math>\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</math>. | ||
+ | |||
+ | However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{075}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | <math>a = \log{x}</math> | ||
+ | |||
+ | <math>b = \log{y}</math> | ||
+ | |||
+ | <math>c = \log{z}</math> | ||
+ | |||
+ | <math>xyz = 10^{81}</math> | ||
+ | |||
+ | <math>\log{xyz} = 81</math> | ||
+ | |||
+ | <math>\log{x} + \log{y} + \log{z} = 81</math> | ||
+ | |||
+ | <math>a + b + c = 81</math> | ||
+ | |||
+ | <math>\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468</math> | ||
+ | |||
+ | <math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math> | ||
+ | |||
+ | <math>a^2 + b^2 + c^2 = 5625 = 75^2</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 | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2010|num-b=4|num-a=6|n=II}} | ||
+ | {{MAA Notice}} |
Revision as of 17:56, 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
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.