2013 AMC 12A Problems/Problem 21
Revision as of 02:30, 7 February 2013 by 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...")
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: