Difference between revisions of "2010 AIME II Problems/Problem 5"
Tempaccount (talk | contribs) (Adding problem section) |
Tempaccount (talk | contribs) (Remove extra problem section) |
||
Line 1: | Line 1: | ||
− | |||
− | |||
== 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>. |
Revision as of 16:03, 9 August 2018
Contents
[hide]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
~MathleteMA
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.