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

(→‎Solution 0)
(→‎Solution 0)
 
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==Solution 0==
 
==Solution 0==
−
First, let’s realize the rule that <math>\log{a}{b}=\log{a}+\log{b}</math>. Using this rule with each of the three equations given, 2 equations at a time, we get:
+
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}{z^2}} = \frac{1}{2}+\frac{1}{3}= \frac{5}{6}</math>
Line 23: Line 23:
 
We can convert the equations above and setting them variables to:
 
We can convert the equations above and setting them variables to:
  
−
a = <math>\log_2{x^2}=\log_2{1}-\frac{7}{12}</math>
+
<math>a = \log_2{x^2}=\log_2{1}-\frac{7}{12}</math>
  
−
b = <math>\log_2{y^2}=\log_2{1}-\frac{5}{6}</math>
+
<math>b = \log_2{y^2}=\log_2{1}-\frac{3}{4}</math>
  
−
c = <math>\log_2{z^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>
 
Then, using the first rule <math>a^2bc = 4(\log_2{1})-\frac{11}{4}</math>
Line 36: Line 36:
 
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>
 
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{5}{6}</math>, we get:  
+
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{5}{6}</math>
+
<math>2\log_2{y} = \log_2{1}-\frac{3}{4}</math>
  
 
Which gives:
 
Which gives:

Latest revision as of 22:35, 14 May 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 $m+n$.

Video Solution! Quick, Easy, Fast, Simple!

https://youtu.be/-VDnZ_iWnBM

~MathKatana

Solution 0

First, let’s realize the rule that $\log{a}{b}=\log{a}+\log{b}$. If we add two equations at a time, and use this rule, 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:

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

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

$c = \log_2{z^2}=\log_2{1}-\frac{5}{6}$

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 = $\log_2{y^2}=\log_2{1}-\frac{3}{4}$, we get:

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

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)

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

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