Difference between revisions of "2024 AIME II Problems/Problem 4"
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Now, we can solve to get <math>a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}</math>. Plugging these values in, we obtain <math>|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}</math>. ~akliu | Now, we can solve to get <math>a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}</math>. Plugging these values in, we obtain <math>|4a + 3b + 2c| = \frac{25}{8} \implies \boxed{033}</math>. ~akliu | ||
+ | ==Solution 2== | ||
+ | <math>\log_2(y/xz) + \log_2(z/xy) = \log_2(1/x^2) = -2\log_2(x) = \frac{7}{12}</math> | ||
+ | |||
+ | <math>\log_2(x/yz) + \log_2(z/xy) = \log_2(1/y^2) = -2\log_2(y) = \frac{3}{4}</math> | ||
+ | |||
+ | <math>\log_2(x/yz) + \log_2(y/xz) = \log_2(1/z^2) = -2\log_2(z) = \frac{5}{6}</math> | ||
+ | |||
+ | <math>\log_2(x) = -\frac{7}{24}</math> | ||
+ | |||
+ | <math>\log_2(y) = -\frac{3}{8}</math> | ||
+ | |||
+ | <math>\log_2(z) = -\frac{5}{12}</math> | ||
+ | |||
+ | <math>4\log_2(x) + 3\log_2(y) + 2\log_2(z) = -25/8</math> | ||
+ | |||
+ | <math>25 + 8 = \boxed{33}</math> | ||
==See also== | ==See also== |
Revision as of 03:20, 9 February 2024
Contents
Problem
Let and
be positive real numbers that satisfy the following system of equations:
Then the value of
is
where
and
are relatively prime positive integers. Find
.
Solution 1
Denote ,
, and
.
Then, we have:
Now, we can solve to get . Plugging these values in, we obtain
. ~akliu
Solution 2
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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