Difference between revisions of "2019 AMC 12A Problems/Problem 23"
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Therefore, <math>a_n = n^{\frac{1}{\log_{7}(2)}} = n \, \heartsuit \, 2</math> for all <math>n \geq 3</math>, completing the induction. | Therefore, <math>a_n = n^{\frac{1}{\log_{7}(2)}} = n \, \heartsuit \, 2</math> for all <math>n \geq 3</math>, completing the induction. | ||
− | We have <math>a_{2019} = 2019^{\log_{2}(7)}</math>. Taking log base 2019 of both sides gives us <math>{\log_{2019}(a_{2019})} = {\log_{2}(7)}</math>. Then, by changing to base 7 and after cancellation, we arrive at <math>{\log_{7}(a_{2019})} = {\log_{2}(2019)}</math>. Because <math>2^{11} = 2048</math> and <math>2^{10} = 1024</math>, our answer is <math>\boxed{\textbf{(D)}11}</math> | + | We have <math>a_{2019} = 2019^{\log_{2}(7)}</math>. Taking log base 2019 of both sides gives us <math>{\log_{2019}(a_{2019})} = {\log_{2}(7)}</math>. Then, by changing to base 7 and after cancellation, we arrive at <math>{\log_{7}(a_{2019})} = {\log_{2}(2019)}</math>. Because <math>2^{11} = 2048</math> and <math>2^{10} = 1024</math>, our answer is <math>\boxed{\textbf{(D) } 11}</math>. |
==See Also== | ==See Also== |
Revision as of 20:32, 10 February 2019
Contents
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 . Changing bases to
, this becomes
, so
, where the last equality comes from the logarithmic chain rule. 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.
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 log base 2019 of both sides gives us
. Then, by changing to base 7 and after cancellation, we arrive at
. Because
and
, our answer is
.
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.