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

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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
 
With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
  
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== See also ==
 
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{{AIME box|year=1983|before=First Question|num-a=2}}
* [[1983 AIME Problems/Problem 2|Next Problem]]
 
* [[1983 AIME Problems|Back to Exam]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 17:34, 21 March 2007

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 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 $\log_zw=60$.

See also

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