AoPSWiki

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

Personal tools

Search

Toolbox

1983 AIME Problems/Problem 1

From AoPSWiki

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 $\log_zw$ .

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

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,

$\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\\log_y ...$

Therefore, $\frac {\log z}{\logw} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$ .

Hence, $\log_z w = \boxed{060}$ .

See also

1983 AIME (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_1"

Category: Intermediate Algebra Problems

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us