Difference between revisions of "2024 AIME II Problems/Problem 4"
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<cmath>\log_2\left({x \over yz}\right) = {1 \over 2}</cmath><cmath>\log_2\left({y \over xz}\right) = {1 \over 3}</cmath><cmath>\log_2\left({z \over xy}\right) = {1 \over 4}</cmath> | <cmath>\log_2\left({x \over yz}\right) = {1 \over 2}</cmath><cmath>\log_2\left({y \over xz}\right) = {1 \over 3}</cmath><cmath>\log_2\left({z \over xy}\right) = {1 \over 4}</cmath> | ||
Then the value of <math>\left|\log_2(x^4y^3z^2)\right|</math> is <math>\tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Then the value of <math>\left|\log_2(x^4y^3z^2)\right|</math> is <math>\tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==Video Solution! Quick, Easy, Fast, Simple!== | ||
+ | https://youtu.be/-VDnZ_iWnBM | ||
+ | |||
+ | ~MathKatana | ||
+ | |||
+ | ==Solution 0== | ||
+ | First, let’s realize the rule that <math>\log{a}{b}=\log{a}+\log{b}</math>. If we add two equations at a time, and use this rule, we get: | ||
+ | |||
+ | <math>\log_2{\frac{1}{z^2}} = \frac{1}{2}+\frac{1}{3}= \frac{5}{6}</math> | ||
+ | |||
+ | <math>\log_2{\frac{1}{x^2}} = \frac{1}{3}+\frac{1}{4}= \frac{7}{12}</math> | ||
+ | |||
+ | <math>\log_2{\frac{1}{y^2}} = \frac{1}{2}+\frac{1}{4}= \frac{3}{4}</math> | ||
+ | |||
+ | Now we look into the rule <math>\log{\frac{b}{c}}=\log{b}-\log{c}</math> | ||
+ | |||
+ | We can convert the equations above and setting them variables to: | ||
+ | |||
+ | <math>a = \log_2{x^2}=\log_2{1}-\frac{7}{12}</math> | ||
+ | |||
+ | <math>b = \log_2{y^2}=\log_2{1}-\frac{3}{4}</math> | ||
+ | |||
+ | <math>c = \log_2{z^2}=\log_2{1}-\frac{5}{6}</math> | ||
+ | |||
+ | Then, using the first rule <math>a^2bc = 4(\log_2{1})-\frac{11}{4}</math> | ||
+ | Make sure you see why there is <math>a^2</math>! We are trying to get the absolute value equation. | ||
+ | |||
+ | We are still missing one y in our <math>\log_2{x^4y^2z^2}</math> | ||
+ | |||
+ | How do we get this? We introduce yet another rule(I know, rules are annoying, but useful and easy to memorize once you derive them yourself), <math>\log{b^n}=n\log{b}</math> | ||
+ | |||
+ | Using this in our equation b = <math>\log_2{y^2}=\log_2{1}-\frac{3}{4}</math>, we get: | ||
+ | |||
+ | <math>2\log_2{y} = \log_2{1}-\frac{3}{4}</math> | ||
+ | |||
+ | Which gives: | ||
+ | |||
+ | <math>\log_2{y}=\frac{\log_2{1}}{2}-\frac{3}{8}</math> | ||
+ | |||
+ | Now, using the first rule again, we combine this with <math>\log_2{x^4y^2z^2}</math> to get our desired equation! We yield: | ||
+ | |||
+ | <math>\log_2{x^4y^3z^2}</math> = <math>\frac{9\log_2{1}}{2}-\frac{25}{8}</math> | ||
+ | |||
+ | Then, we feel sad because we don’t know what <math>\log_2{1}</math> is. But then we realize, 2 to the power of 0 is 1! So we can just cancel out everything before the <math>-\frac{25}{8}</math>. | ||
+ | |||
+ | Therefore, <math>\log_2{x^4y^3z^2}</math> is <math>-\frac{25}{8}</math>. After absolute value, it is just <math>\frac{25}{8}</math>. Summing m and n, we obtain <math>\boxed{033}</math> | ||
+ | |||
+ | ~MathKatana (This was written by a 6th grader, any issues please report!) | ||
==Solution 1== | ==Solution 1== | ||
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{{AIME box|year=2024|num-b=3|num-a=5|n=II}} | {{AIME box|year=2024|num-b=3|num-a=5|n=II}} | ||
− | [[Category:]] | + | [[Category:Intermediate Algebra Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:35, 14 May 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
.
Video Solution! Quick, Easy, Fast, Simple!
~MathKatana
Solution 0
First, let’s realize the rule that . If we add two equations at a time, and use this rule, we get:
Now we look into the rule
We can convert the equations above and setting them variables to:
Then, using the first rule
Make sure you see why there is
! We are trying to get the absolute value equation.
We are still missing one y in our
How do we get this? We introduce yet another rule(I know, rules are annoying, but useful and easy to memorize once you derive them yourself),
Using this in our equation b = , we get:
Which gives:
Now, using the first rule again, we combine this with to get our desired equation! We yield:
=
Then, we feel sad because we don’t know what is. But then we realize, 2 to the power of 0 is 1! So we can just cancel out everything before the
.
Therefore, is
. After absolute value, it is just
. Summing m and n, we obtain
~MathKatana (This was written by a 6th grader, any issues please report!)
Solution 1
Denote ,
, and
.
Then, we have:
Now, we can solve to get . Plugging these values in, we obtain
. ~akliu
Solution 2
~Callisto531
Solution 3
Adding all three equations, . Subtracting this from every equation, we have:
Our desired quantity is the absolute value of
, so our answer is
.
~Spoirvfimidf
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.