Difference between revisions of "2003 AIME I Problems/Problem 4"
(→Solution 3) |
(→Solution 3) |
||
Line 52: | Line 52: | ||
From here it is obvious that <math>\boxed{n=012}</math>. | From here it is obvious that <math>\boxed{n=012}</math>. | ||
+ | |||
+ | |||
+ | ~yofro | ||
== See also == | == See also == |
Latest revision as of 20:59, 22 October 2020
Problem
Given that and that find
Solution 1
Using the properties of logarithms, we can simplify the first equation to . Therefore,
Now, manipulate the second equation.
By the Pythagorean identities, , and we can substitute the value for from . .
Solution 2
Examining the first equation, we simplify as the following:
With this in mind, examining the second equation, we may simplify as the following (utilizing logarithm properties):
From here, we may divide both sides by and then proceed with the change-of-base logarithm property:
Thus, exponentiating both sides results in . Squaring both sides gives us
Via the Pythagorean Identity, and is simply , via substitution. Thus, substituting these results into the current equation:
Using simple cross-multiplication techniques, we have , and thus . ~ nikenissan
Solution 3
By the first equation, we get that . We can let , . Thus . By the identity , we get that . Solving this, we get . So we have
From here it is obvious that .
~yofro
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.