2019 AMC 12A Problems/Problem 23
Problem
Define binary operations and
by
for all real numbers
and
for which these expressions are defined. The sequence
is defined recursively by
and
for all integers
. To the nearest integer, what is
?
Solution 1
By definition, the recursion becomes . By the change of base formula, this reduces to
. Thus, we have
. Thus, for each positive integer
, the value of
must be some constant value
.
We now compute from
. It is given that
, so
.
Now, we must have . At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.
We conclude that , or choice
.
Solution 2
Using the recursive definition, or
where
and
. Using logarithm rules, we can remove the exponent of the 3 so that
. Therefore,
, which is
.
We claim that for all
. We can prove this through induction.
Clearly, the base case where holds.
This can be simplified as .
Applying the diamond operation, we can simplify where
. By using logarithm rules to remove the exponent of
and after cancelling,
.
Therefore, for all
, completing the induction.
We have . Taking
of both sides gives us
. Then, by changing to base
and after cancellation, we arrive at
. Because
and
, our answer is
.
Solution 3
We are given that
Since we are asked to find
, we directly apply
Using the property that
Now using the property that
Once again applying the first property yields
Rearranging the expression,
Now expressing in a similar expression as
,
Because of the fact that , we can cancel out the terms to get
Using the Chain Rule for Logarithm, , yields
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.