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

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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
 
With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
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----
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== Alternative Solution ==
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Applying Change of Base Formula:
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<cmath> \log_x w = 24 \implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} </cmath>
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<cmath> \log_y w = 40 \implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} </cmath>
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<cmath> \log_{xyz} w = 12 \implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} </cmath>
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Therefore, <math> \frac {\log z}{\log
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w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>.
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Hence, <math> \log_z w = 60</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:43, 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, $\log_z w = 60$.

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