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

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== Problem ==
 
== Problem ==
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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_xw=24</math>, <math>\log_y w = 40</math>, and <math>\log_{xyz}w=12</math>. Find <math>\log_zw</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>.
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== Solutions ==
  
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__TOC__
 
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== Solution ==
 
 
=== Solution 1 ===
 
=== Solution 1 ===
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The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential [[expression]]s.  
 
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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Hence, <math> \log_z w = \boxed{060}</math>.
 
Hence, <math> \log_z w = \boxed{060}</math>.
  
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{{alternate solutions}}
  
 
== See Also ==
 
== See Also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 04:37, 8 August 2014

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

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*}$ (Error compiling LaTeX. Unknown error_msg)

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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