Difference between revisions of "1983 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
− | Let <math>x</math>,<math>y</math>, and <math>z</math> all exceed <math>1</math>, and let <math>w</math> be a [[positive number]] such that <math>\ | + | Let <math>x</math>, <math>y</math>, and <math>z</math> all exceed <math>1</math>, and let <math>w</math> be a [[positive number]] such that <math>\log_x w = 24</math>, <math>\log_y w = 40</math>, and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>. |
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+ | == Solutions == | ||
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=== Solution 1 === | === Solution 1 === | ||
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The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential [[expression]]s. | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential [[expression]]s. | ||
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Hence, <math> \log_z w = \boxed{060}</math>. | Hence, <math> \log_z w = \boxed{060}</math>. | ||
+ | {{alternate solutions}} | ||
== See Also == | == See Also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 04:37, 8 August 2014
Contents
[hide]Problem
Let , , and all exceed , and let be a positive number such that , , and . Find .
Solutions
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, .
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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.