Difference between revisions of "2010 AIME II Problems/Problem 5"
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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>. | 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{ | + | 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>. |
== 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}} | ||
Revision as of 22:26, 9 November 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 
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 | ||
