2024 AIME II Problems/Problem 4

Problem

Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations: \[\log_2\left({x \over yz}\right) = {1 \over 2}\]\[\log_2\left({y \over xz}\right) = {1 \over 3}\]\[\log_2\left({z \over xy}\right) = {1 \over 4}\] Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Denote $\log_2(x) = a$, $\log_2(y) = b$, and $\log_2(z) = c$.

Then, we have: $a-b-c = \frac{1}{2}$ $-a+b-c = \frac{1}{3}$ $-a-b+c = \frac{1}{4}$

Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$. Plugging these values in, we obtain $|4a + 3b + 2c|  = \frac{25}{8} \implies \boxed{033}$. ~akliu

Solution 2

$\log_2(y/xz) + \log_2(z/xy) = \log_2(1/x^2) = -2\log_2(x) = \frac{7}{12}$

$\log_2(x/yz) + \log_2(z/xy) = \log_2(1/y^2) = -2\log_2(y) = \frac{3}{4}$

$\log_2(x/yz) + \log_2(y/xz) = \log_2(1/z^2) = -2\log_2(z) = \frac{5}{6}$

$\log_2(x) = -\frac{7}{24}$

$\log_2(y) = -\frac{3}{8}$

$\log_2(z) = -\frac{5}{12}$

$4\log_2(x) + 3\log_2(y) + 2\log_2(z) = -25/8$

$25 + 8 = \boxed{033}$

~Callisto531

Solution 3

Adding all three equations, $\log_2(\frac{1}{xyz}) = \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{13}{12}$. Subtracting this from every equation, we have: \[2\log_2x = -\frac{7}{12},\] \[2\log_2y = -\frac{3}{4},\] \[2\log_2z = -\frac{5}{6}\] Our desired quantity is the absolute value of $4\log_2x+3\log_2y+2\log_2z = 2(\frac{7}{12})+3/2(\frac{3}{4})+\frac{5}{6} = \frac{25}{8}$, so our answer is $25+8 = \boxed{033}$. ~Spoirvfimidf

Video Solution

https://youtu.be/SUie2Jlo-pg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2024 AIME II (ProblemsAnswer KeyResources)
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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