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

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== Problem ==
 
== Problem ==
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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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>.
  
 
== Solutions ==
 
== Solutions ==
  
 
=== Solution 1 ===
 
=== Solution 1 ===
The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential [[expression]]s.  
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The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
  
 
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>.
 
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>.

Revision as of 18:00, 15 February 2019

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 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 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 (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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