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

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=== Solution 5 ===
 
=== Solution 5 ===
  
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If we convert all of the equations into exponential form, we receive <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The last equation can also be written as <math>x^{12}y^{12}z^{12}=w</math>. Also note that <math>x^{24}y^{40}= w^{2}</math>. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. However, <math>x^{12}y^{12}z^{12}=w</math>. Thus, after <math>z^{12}= y^{8}</math>. Thus <math>\log_z w</math>= 12<math>\log_y w</math>/{8}<math> = \boxed{060}</math>.
+
If we convert all of the equations into exponential form, we receive <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The last equation can also be written as <math>x^{12}y^{12}z^{12}=w</math>. Also note that <math>x^{24}y^{40}= w^{2}</math>. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. However, <math>x^{12}y^{12}z^{12}=w</math>. Thus, after <math>z^{12}= y^{8}</math>. Thus <math>\log_z w</math>= 12<math>\log_y w</math>/8<math> = \boxed{060}</math>.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 15:38, 8 December 2020

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 $\log_z w$.

Solution

Solution 1

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

$x^{24}=w$, $y^{40}=w$, 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 containing $z$ is $(xyz)^{12}=w$. It now becomes clear that 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 $x^{12}=w^{\frac{1}{2}}$. Raising both sides of $y^{40}=w$ to the $\frac{12}{40}$th power gives $y^{12}=w^{\frac{3}{10}}$.

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 $w^{1/2}w^{3/10}z^{12}=w$. Simplifying, 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, $\log_z w = \boxed{060}$.

Solution 4

Since $\log_a b = \frac{1}{\log_b a}$, the given conditions can be rewritten as $\log_w x = \frac{1}{24}$, $\log_w y = \frac{1}{40}$, and $\log_w xyz = \frac{1}{12}$. Since $\log_a \frac{b}{c} = \log_a b - \log_a c$, $\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}$. Therefore, $\log_z w = \boxed{060}$.

Solution 5

If we convert all of the equations into exponential form, we receive $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. The last equation can also be written as $x^{12}y^{12}z^{12}=w$. Also note that $x^{24}y^{40}= w^{2}$. Taking the square root of this, we find that $x^{12}y^{20}=w$. However, $x^{12}y^{12}z^{12}=w$. Thus, after $z^{12}= y^{8}$. Thus $\log_z w$= 12$\log_y w$/8$= \boxed{060}$.

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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