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Difference between revisions of "2024 AIME II Problems/Problem 4"
Line 14: Line 14:
 
<math>-a-b+c = \frac{1}{4}</math>
 
<math>-a-b+c = \frac{1}{4}</math>
  
−
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>\mid 4a + 3b + 2c \mid = \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

Revision as of 19:03, 8 February 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 ${m \over n}$ 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

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