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 0

First, let’s realize the rule that $\log{a}{b}=\log{a}+\log{b}$ . Using this rule with each of the three equations given, 2 equations at a time, we get:

$\log_2{\frac{1}{z^2}} = \frac{1}{2}+\frac{1}{3}= \frac{5}{6}$

$\log_2{\frac{1}{x^2}} = \frac{1}{3}+\frac{1}{4}= \frac{7}{12}$

$\log_2{\frac{1}{y^2}} = \frac{1}{2}+\frac{1}{4}= \frac{3}{4}$

Now we look into the rule $\log{\frac{b}{c}}=\log{b}-\log{c}$

We can convert the equations above and setting them variables to:

a = $\log_2{x^2}=\log_2{1}-\frac{7}{12}$

b = $\log_2{y^2}=\log_2{1}-\frac{5}{6}$

c = $\log_2{z^2}=\log_2{1}-\frac{3}{4}$

Then, using the first rule $a^2bc = 4(\log_2{1})-\frac{11}{4}$ Make sure you see why there is $a^2$ ! We are trying to get the absolute value equation.

We are still missing one y in our $\log_2{x^4y^2z^2}$

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), $\log{b^n}=n\log{b}$

Using this in our equation b = $\log_2{y^2}=\log_2{1}-\frac{5}{6}$ , we get:

$2\log_2{y} = \log_2{1}-\frac{5}{6}$

Which gives:

$\log_2{y}=\frac{\log_2{1}}{2}-\frac{3}{8}$

Now, using the first rule again, we combine this with $\log_2{x^4y^2z^2}$ to get our desired equation! We yield:

$\log_2{x^4y^3z^2}$ = $\frac{9\log_2{1}}{2}-\frac{25}{8}$

Then, we feel sad because we don’t know what $\log_2{1}$ is. But then we realize, 2 to the power of 0 is 1! So we can just cancel out everything before the $-\frac{25}{8}$ .

Therefore, $\log_2{x^4y^3z^2}$ is $-\frac{25}{8}$ . After absolute value, it is just $\frac{25}{8}$ . Summing m and n, we obtain $\boxed{033}$

~MathKatana (This was written by a 6th grader, any issues please report!)

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)

2024 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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