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

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== Solution ==
 
== Solution ==
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The logarithmic notation doesn't tell us much, so we'll first convert everything to exponents.
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<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>.
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<math>x^{120}=w^5</math>, <math>y^{120}=w^3</math>, and <math>(xyz)^{120}=w^{10}</math>.
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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
  
 
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Revision as of 23:49, 23 July 2006

Problem

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

Solution

The logarithmic notation doesn't tell us much, so we'll first convert everything to exponents.

$x^{24}=w$, $y^{40}=w$, 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 $\log_zw=60$.