Difference between revisions of "1983 AIME Problems/Problem 1"

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Taking the square root of the equation <math>x^{24}=w</math> results in <math>x^{12}=w^{1/2}</math>. Taking the <math>12/40</math> root of <math>y^{40}=w</math> equates to <math>y^{12}=w^{3/10}</math>.
 
Taking the square root of the equation <math>x^{24}=w</math> results in <math>x^{12}=w^{1/2}</math>. Taking the <math>12/40</math> root of <math>y^{40}=w</math> equates to <math>y^{12}=w^{3/10}</math>.
  
Going back to <math>(xyz)^{12}=w</math>, we can substitute the <math>x^{12}</math> and <math>y^{12}</math> with <math>w^{1/2}</math> and <math>w^{3/10}</math>, respectively. We now have <math>w^{1/2}<cmath>w^{3/10}</cmath>z^{12}=w</math>. Simplify we get <math>z^{60}=w</math>.
+
Going back to <math>(xyz)^{12}=w</math>, we can substitute the <math>x^{12}</math> and <math>y^{12}</math> with <math>w^{1/2}</math> and <math>w^{3/10}</math>, respectively. We now have <math>w^{1/2}</math>*<math>w^{3/10}</math>*<math>z^{12}=w</math>. Simplify we get <math>z^{60}=w</math>.
 
So our answer is \boxed{060}<math>
 
So our answer is \boxed{060}<math>
  

Revision as of 18:52, 5 August 2016

Problem

Let $x$, $y$, and $z$ all exceed $1$, and let $w$ be a positive number such that $\log_x w = 24$, $\log_y w = 40$, and $\log_{xyz} w = 12$. Find $\log_z w$.

Solutions

Solution 1

The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential expressions.

$x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. If we now convert everything to a power of $120$, it will be easy to isolate $z$ and $w$.

$x^{120}=w^5$, $y^{120}=w^3$, and $(xyz)^{120}=w^{10}$.

With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\log_zw=\boxed{060}$.

Solution 2

First we'll convert everything to exponential form. $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. The only expression with z is $(xyz)^{12}=w$. It now becomes clear one way to find $\log_z w$ is to find what $x^{12}$ and $y^{12}$ are in terms of $w$.

Taking the square root of the equation $x^{24}=w$ results in $x^{12}=w^{1/2}$. Taking the $12/40$ root of $y^{40}=w$ equates to $y^{12}=w^{3/10}$.

Going back to $(xyz)^{12}=w$, we can substitute the $x^{12}$ and $y^{12}$ with $w^{1/2}$ and $w^{3/10}$, respectively. We now have $w^{1/2}$*$w^{3/10}$*$z^{12}=w$. Simplify we get $z^{60}=w$. So our answer is \boxed{060}$=== Solution 3 === Applying the change of base formula, <cmath>logxw=24logwlogx=24logxlogw=124logyw=40logwlogy=40logylogw=140logxyzw=12logwlogxyz=12logx+logy+logzlogw=112</cmath> Therefore,$ (Error compiling LaTeX. Unknown error_msg) \frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$.

Hence,$ (Error compiling LaTeX. Unknown error_msg) \log_z w = \boxed{060}$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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