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

## Solution

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 $f(x-1)$: $f_{0}(x) \approx \log x$

And a better approximation, by plugging in our first approximation for $f(x-1)$ in our original definition for $f(x)$: $f_{1}(x) \approx \log(x + \log(x-1))$

And an even better approximation: $f_{2}(x) \approx \log(x + \log(x-1 + \log(x-2)))$

Continuing this pattern, obviously, will eventually terminate at $f_{x-1}(x)$, in other words our original definition of $f(x)$.

However, at $x = 2013$, going further than $f_{1}(x)$ will not distinguish between our answer choices. $\log(2012 + \log(2011))$ is nearly indistinguishable from $\log(2012)$.

So we take $f_{1}(x)$ 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: $\log 2016 < \log(2013 + \log 2012) < \log(2017)$ $(\log 2016, \log 2017)$