Difference between revisions of "2024 AIME II Problems/Problem 4"

Revision as of 17:08, 5 March 2024

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)

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 {{AIME box|year=2024|num-b=3|num-a=5|n=II}}
-[[Category:]]
+[[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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