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 -_-) |
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==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 20:33, 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.