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}x)= 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{75}</math>. | 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}x)= 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{75}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2010|num-b=4|num-a=6|n=II}} | ||
Revision as of 16:11, 3 April 2010
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 $\(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2$ (Error compiling LaTeX. Unknown error_msg) is equal to 
. However, we want to find , so we take the square root of 
, or 
.
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 | ||
