AoPSWiki

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

Personal tools

Log in

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

Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.

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