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Difference between revisions of "1983 AIME Problems/Problem 1"
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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>.
 
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}
  
 
=== Solution 3 ===
 
=== Solution 3 ===
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\log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\
 
\log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\
 
\log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}</cmath>
 
\log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}</cmath>
−
Therefore, </math> \frac {\log z}{\log
+
Therefore, <math> \frac {\log z}{\log
−
w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}<math>.  
+
w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>.  
  
−
Hence, </math> \log_z w = \boxed{060}$.
+
Hence, <math> \log_z w = \boxed{060}</math>.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

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, \begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ \log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\ \log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*} Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$.

Hence, $\log_z w = \boxed{060}$.

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 (Problems • Answer Key • Resources)
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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