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

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Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w = 24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>.
 
Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w = 24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>.
  
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== Solution ==
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== Solution 1 ==
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=== Solution 1 ===
 
 
The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
 
The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
  
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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>.
 
With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>.
  
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=== Solution 2 ===
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== Solution 2 ==
 
First we'll convert everything to exponential form.
 
First we'll convert everything to exponential form.
 
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The only expression containing <math>z</math> is <math>(xyz)^{12}=w</math>. It now becomes clear that one way to find <math>\log_z w</math> is to find what <math>x^{12}</math> and <math>y^{12}</math> are in terms of <math>w</math>.  
 
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The only expression containing <math>z</math> is <math>(xyz)^{12}=w</math>. It now becomes clear that one way to find <math>\log_z w</math> is to find what <math>x^{12}</math> and <math>y^{12}</math> are in terms of <math>w</math>.  
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So our answer is <math>\boxed{060}</math>.
 
So our answer is <math>\boxed{060}</math>.
  
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=== Solution 3 ===
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== Solution 3 ==
 
Applying the change of base formula,
 
Applying the change of base formula,
 
<cmath>\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\
 
<cmath>\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\
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Hence, <math> \log_z w = \boxed{060}</math>.
 
Hence, <math> \log_z w = \boxed{060}</math>.
  
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=== Solution 4 ===
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== Solution 4 ==
 
Since <math>\log_a b = \frac{1}{\log_b a}</math>, the given conditions can be rewritten as <math>\log_w x = \frac{1}{24}</math>, <math>\log_w y = \frac{1}{40}</math>, and <math>\log_w xyz = \frac{1}{12}</math>. Since <math>\log_a \frac{b}{c} = \log_a b - \log_a c</math>,  <math>\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}</math>. Therefore, <math>\log_z w = \boxed{060}</math>.
 
Since <math>\log_a b = \frac{1}{\log_b a}</math>, the given conditions can be rewritten as <math>\log_w x = \frac{1}{24}</math>, <math>\log_w y = \frac{1}{40}</math>, and <math>\log_w xyz = \frac{1}{12}</math>. Since <math>\log_a \frac{b}{c} = \log_a b - \log_a c</math>,  <math>\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}</math>. Therefore, <math>\log_z w = \boxed{060}</math>.
  
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=== Solution 5 ===
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== Solution 5 ==
 +
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 by multiplying the first two equations, we get, <math>x^{24}y^{40}= w^{2}</math>. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. Recall, <math>x^{12}y^{12}z^{12}=w</math>. Thus, <math>z^{12}= y^{8}</math>. Also recall, <math>y^{40}=w</math>. Therefore, <math>z^{60}</math> = <math>y^{40}</math> = <math>w</math>. So, <math>\log_z w</math> = <math>\boxed{060}</math>.
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-Dhillonr25, Bobbob
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== Solution 6 ==
 +
Converting all of the logarithms to exponentials gives <math>x^{24} = w, y^{40} =w,</math> and <math>x^{12}y^{12}z^{12}=w.</math>
 +
Thus, we have <math>y^{40} = x^{24} \Rightarrow z^3=y^2.</math>
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We are looking for <math>\log_z w,</math> which by substitution, is <math>\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.</math>
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 +
~coolmath2017
  
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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}/{8}</math> = \boxed{060}$.
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== Video Solution ==
 +
https://youtu.be/8XjBNtFWWww
  
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{{alternate solutions}}
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~Lucas
  
 
== See Also ==
 
== See Also ==

Latest revision as of 02:14, 17 August 2022

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 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 by multiplying the first two equations, we get, $x^{24}y^{40}= w^{2}$. Taking the square root of this, we find that $x^{12}y^{20}=w$. Recall, $x^{12}y^{12}z^{12}=w$. Thus, $z^{12}= y^{8}$. Also recall, $y^{40}=w$. Therefore, $z^{60}$ = $y^{40}$ = $w$. So, $\log_z w$ = $\boxed{060}$.

-Dhillonr25, Bobbob

Solution 6

Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$

~coolmath2017

Video Solution

https://youtu.be/8XjBNtFWWww

~Lucas

See Also

1983 AIME (Problems • Answer Key • Resources)
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First Question 		Followed by
Problem 2
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All AIME Problems and Solutions

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