1983 AIME Problems/Problem 1
Contents
Problem
Let , and all exceed and let be a positive number such that , and . Find .
Solution
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 containing is . It now becomes clear that one way to find is to find what and are in terms of .
Taking the square root of the equation results in . Raising both sides of to the th power gives .
Going back to , we can substitute the and with and , respectively. We now have . Simplifying, we get . So our answer is .
Solution 3
Applying the change of base formula, Therefore, .
Hence, .
Solution 4
Since , the given conditions can be rewritten as , , and . Since , . Therefore, .
Solution 5
If we convert all of the equations into exponential form, we receive , , and . The last equation can also be written as . Also note that by multiplying the first two equations, we get, . Taking the square root of this, we find that . Recall, . Thus, . Also recall, . Therefore, = = . So, = .
-Dhillonr25, Bobbob
Solution 6
Converting all of the logarithms to exponentials gives and Thus, we have We are looking for which by substitution, is
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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