Difference between revisions of "2016 AIME II Problems/Problem 3"
m (→Problem: ; reformatted for cleanliness) |
m (→Problem: ; because apparently wrapping align* n double dollars makes sense -_-) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system | Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system | ||
| − | \begin{align*} | + | <cmath>\begin{align*} |
\log_2(xyz-3+\log_5 x)&=5,\\ | \log_2(xyz-3+\log_5 x)&=5,\\ | ||
\log_3(xyz-3+\log_5 y)&=4,\\ | \log_3(xyz-3+\log_5 y)&=4,\\ | ||
\log_4(xyz-3+\log_5 z)&=4.\\ | \log_4(xyz-3+\log_5 z)&=4.\\ | ||
| − | \end{align*} | + | \end{align*}</cmath> |
Find the value of <math>|\log_5 x|+|\log_5 y|+|\log_5 z|</math>. | Find the value of <math>|\log_5 x|+|\log_5 y|+|\log_5 z|</math>. | ||
Latest revision as of 19:33, 8 November 2022
Problem
Let 
and 
be real numbers satisfying the system
Find the value of 
.
Solution
First, we get rid of logs by taking powers: 
, 
, and 
. Adding all the equations up and using the 
property, we have 
, so we have 
. Solving for 
by substituting 
for 
in each equation, we get 
, so adding all the absolute values we have 
.
Note: because we know has to be a power of 
, and so it is not hard to test values in the equation in order to achieve desired value for .
See also
| 2016 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
