Difference between revisions of "1983 AIME Problems/Problem 1"
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w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. | w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. | ||
− | Hence, <math> \log_z w = | + | Hence, <math> \log_z w = \boxed{060}</math>. |
== See also == | == See also == |
Revision as of 10:34, 25 April 2008
Problem
Let ,, and all exceed , and let be a positive number such that , , and . Find .
Contents
[hide]Solution
Solution 1
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 .
Solution 2
Applying the change of base formula,
\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*}$ (Error compiling LaTeX. Unknown error_msg)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 |