Difference between revisions of "1983 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
− | Let <math>x</math>, <math>y</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>. |
== Solutions == | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
− | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential | + | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. |
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>. | <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>. |
Revision as of 18:00, 15 February 2019
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 forms.
, , 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
First we'll convert everything to exponential form. , , and . The only expression with z is . It now becomes clear one way to find is to find what and are in terms of .
Taking the square root of the equation results in . Taking the root of equates to .
Going back to , we can substitute the and with and , respectively. We now have **. Simplify we get . So our answer is .
Solution 3
Applying the change of base formula, 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.