Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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<math>f(2013) \approx \log(2013 + \log 2012)</math> | <math>f(2013) \approx \log(2013 + \log 2012)</math> | ||
− | Since <math>1000 < 2012 < 10000</math>, we know <math>3 < log(2012) < 4</math>. This gives us our answer range: | + | Since <math>1000 < 2012 < 10000</math>, we know <math>3 < \log(2012) < 4</math>. This gives us our answer range: |
<math>\log 2016 < \log(2013 + \log 2012) < \log(2017)</math> | <math>\log 2016 < \log(2013 + \log 2012) < \log(2017)</math> | ||
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Suppose <math>A=\log(x)</math>. | Suppose <math>A=\log(x)</math>. | ||
Then <math>\log(2012+ \cdots)=x-2013</math>. | Then <math>\log(2012+ \cdots)=x-2013</math>. | ||
− | So if <math>x>2017</math>, then <math>\log(2012+log(2011+\cdots))>4</math>. | + | So if <math>x>2017</math>, then <math>\log(2012+\log(2011+\cdots))>4</math>. |
− | So <math>2012+log(2011+\cdots)>10000</math>. | + | So <math>2012+\log(2011+\cdots)>10000</math>. |
− | Repeating, we then get <math>2011+log(2010+\cdots)>10^{7988}</math>. | + | Repeating, we then get <math>2011+\log(2010+\cdots)>10^{7988}</math>. |
This is clearly absurd (the RHS continues to grow more than exponentially for each iteration). | This is clearly absurd (the RHS continues to grow more than exponentially for each iteration). | ||
So, <math>x</math> is not greater than <math>2017</math>. | So, <math>x</math> is not greater than <math>2017</math>. |
Revision as of 19:25, 4 February 2019
Contents
[hide]Problem
Consider . Which of the following intervals contains ?
Solution 1
Let and , and from the problem description,
We can reason out an approximation, by ignoring the :
And a better approximation, by plugging in our first approximation for in our original definition for :
And an even better approximation:
Continuing this pattern, obviously, will eventually terminate at , in other words our original definition of .
However, at , going further than will not distinguish between our answer choices. is nearly indistinguishable from .
So we take and plug in.
Since , we know . This gives us our answer range:
Solution 2
Suppose . Then . So if , then . So . Repeating, we then get . This is clearly absurd (the RHS continues to grow more than exponentially for each iteration). So, is not greater than . So . But this leaves only one answer, so we are done.
See Also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 |
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