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== 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>\log_y w = 40</math>, and <math>\log_{xyz}w=12</math>. Find <math>\log_zw</math>.

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== Solution ==
=== Solution 1 ===
The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent [[exponential]] [[expression]]s. 

<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^{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=\boxed{060}</math>.

=== Solution 2 ===
Applying the change of base formula,
<center><math>\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*}</math></center>
Therefore, <math> \frac {\log z}{\log
w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. 

Hence, <math> \log_z w = \boxed{060}</math>.

== See also ==
{{AIME box|year=1983|before=First Question|num-a=2}}

[[Category:Intermediate Algebra Problems]]