# Difference between revisions of "2013 AMC 12A Problems/Problem 21"

Epicwisdom (talk | contribs) (Created page with "Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = log(2)</math> We can reason out an approximation, by ignoring the <math>f(x-1)</math>: <math>f(x) \approx \log x</math...") |
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− | Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = log(2)</math> | + | Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = log(2)</math>, and from the problem description, <math>A = f(2013)</math> |

We can reason out an approximation, by ignoring the <math>f(x-1)</math>: | We can reason out an approximation, by ignoring the <math>f(x-1)</math>: |

## Revision as of 02:49, 7 February 2013

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 our original definition of .

However, at , going further than our second approximation will not distinguish between our answer choices.

So we take our second approximation and plug in.

Since , we know . This gives us our answer range: