1983 AIME Problems/Problem 1

Revision as of 15:34, 13 March 2015 by Mathgeek2006 (talk | contribs) (Solution 2)

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

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.

See Also

1983 AIME (ProblemsAnswer KeyResources)
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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