1983 AIME Problems/Problem 1

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, $\log_z w = 60$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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1983 AIME Problems/Problem 1

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Alternative Solution

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