1983 AIME Problems/Problem 1
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
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 |
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