1983 AIME Problems/Problem 1

Problem

Let $x$ , $y$ , and $z$ all exceed $1$ , and let $w$ be a positive number such that $\log_x w = 24$ , $\log_y w = 40$ , and $\log_{xyz} w = 12$ . Find $\log_z w$ .

Solutions

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

First we'll convert everything to exponential form. $x^{24}=w$ , $y^{40}=w$ , and $(xyz)^{12}=w$ . The only expression with z is $(xyz)^{12}=w$ . It now becomes clear one way to find $\log_z w$ is to find what $x^{12}$ and $y^{12}$ are in terms of $w$ .

Taking the square root of the equation $x^{24}=w$ results in $x^{12}=w^{1/2}$ . Taking the $12/40$ root of $y^{40}=w$ equates to $y^{12}=w^{3/10}$ .

Going back to $(xyz)^{12}=w$ , we can substitute the $x^{12}$ and $y^{12}$ with $w^{1/2}$ and $w^{3/10}$ , respectively. We now have $w^{1/2}$ * $w^{3/10}$ * $z^{12}=w$ . Simplify we get $z^{60}=w$ . So our answer is $\boxed{060}$ .

Solution 3

Applying the change of base formula, $\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ \log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\ \log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}$ Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$ .

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

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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