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

(Alternative Solution)
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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_xw=24</math>, <math>\displaystyle \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_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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__TOC__
 
== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
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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<math>x^{120}=w^5</math>, <math>y^{120}=w^3</math>, and <math>(xyz)^{120}=w^{10}</math>.
 
<math>x^{120}=w^5</math>, <math>y^{120}=w^3</math>, and <math>(xyz)^{120}=w^{10}</math>.
  
With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
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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 ===
 
+
Applying the change of base formula,
== Alternative Solution ==
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<center><math>\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \
Applying Change of Base Formula:
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\log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \
<cmath> \log_x w = 24 \implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} </cmath>
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\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*}</math></center>
 
 
<cmath> \log_y w = 40 \implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} </cmath>
 
 
 
<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} </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 = 60</math>.
 
Hence, <math> \log_z w = 60</math>.
 
 
--[[User:Luimichael|Luimichael]] 22:46, 19 November 2007 (EST)
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1983|before=First Question|num-a=2}}
 
{{AIME box|year=1983|before=First Question|num-a=2}}
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 10:33, 25 April 2008

Problem

Let $x$,$y$, and $z$ all exceed $1$, and let $w$ be a positive number such that $\log_xw=24$, $\log_y w = 40$, and $\log_{xyz}w=12$. Find $\log_zw$.

Solution

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, $\log_z w = 60$.

See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions