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

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Hence, <math> \log_z w = 60</math>.
 
Hence, <math> \log_z w = 60</math>.
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--[[User:Luimichael|Luimichael]] 22:46, 19 November 2007 (EST)
  
 
== See also ==
 
== See also ==

Revision as of 22:46, 19 November 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 $\log_zw=60$.

Alternative Solution

Applying Change of Base Formula: \[\log_x w = 24 \implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24}\]

\[\log_y w = 40 \implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40}\]

\[\log_{xyz} w = 12 \implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12}\] Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$.


Hence, $\log_z w = 60$.


--Luimichael 22:46, 19 November 2007 (EST)

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
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