Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(+)
Line 13: Line 13:
 
== 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 13:13, 6 May 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 (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&oldid=14888"