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>. | ||
+ | |||
+ | ---- | ||
+ | |||
+ | == Alternative Solution == | ||
+ | Applying Change of Base Formula: | ||
+ | <cmath> \log_x w = 24 \implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} </cmath> | ||
+ | |||
+ | <cmath> \log_y w = 40 \implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} </cmath> | ||
+ | |||
+ | <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> | ||
+ | Therefore, <math> \frac {\log z}{\log | ||
+ | w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. | ||
+ | |||
+ | |||
+ | Hence, <math> \log_z w = 60</math>. | ||
== See also == | == See also == |
Revision as of 22:43, 19 November 2007
Contents
[hide]Problem
Let ,, and all exceed , and let be a positive number such that , , and . Find .
Solution
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential expressions.
, , and . If we now convert everything to a power of , it will be easy to isolate and .
, , and .
With some substitution, we get and .
Alternative Solution
Applying Change of Base Formula:
Therefore, .
Hence, .
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 |