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

Revision as of 01:49, 7 February 2013

Let $f(x) = \log(x + f(x-1))$ and $f(2) = log(2)$ , and from the problem description, $A = f(2013)$

We can reason out an approximation, by ignoring the $f(x-1)$ :

$f(x) \approx \log x$

And a better approximation, by plugging in our first approximation for $f(x-1)$ in our original definition for $f(x)$ :

$f(x) \approx \log(x + \log(x-1))$

And an even better approximation:

$f(x) \approx \log(x + \log(x-1 + \log(x-2)))$

Continuing this pattern, obviously, will eventually terminate at our original definition of $f(x)$ .

However, at $x = 2013$ , going further than our second approximation will not distinguish between our answer choices.

So we take our second approximation and plug in.

$f(2013) \approx \log(2013 + \log 2012)$

Since $1000 < 2012 < 10000$ , we know $3 < log(2012) < 4$ . This gives us our answer range:

$\log 2016 < \log(2013 + \log 2012) < \log(2017)$

$(\log 2016, \log 2017)$

Revision as of 01:30, 7 February 2013 (view source) 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...")		Revision as of 01:49, 7 February 2013 (view source) Epicwisdom (talk \| contribs) Newer edit →
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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>: