# Difference between revisions of "2009 AMC 12A Problems/Problem 24"

## Problem

The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n\ge1$. Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A$. What is the largest integer $k$ such that

$$\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}$$

is defined?

$\textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013$

## Solution

We just look at the last three logarithms for the moment, and use the fact that $\log_2 T(k) = T(k - 1)$. We wish to find:

$$\log_2\log_2\log_2\left({T(2009)^{\left({T(2009)}^{T(2009)}\right)}}\right)$$ $$= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009)))$$ $$= \log_2(T(2009)\log_2(T(2009)T(2008)))$$ $$= \log_2(T(2009)(T(2008) + T(2007)))$$

Now we realize that $T(n - 1)$ is much smaller than $T(n)$. So we approximate this, remembering we have rounded down, as:

$$\log_2(T(2009)) = T(2008)$$

We have used $3$ logarithms so far. Applying $2007$ more to the left of our expression, we get $T(1) = 2$. Then we can apply the logarithm $2$ more times, until we get to $0$. So our answer is approximately $3 + 2007 + 2 = 2012$. But we rounded down, so that means that after $2012$ logarithms we get a number slightly greater than $0$, so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied $1$ more time since $2012 + 1 = 2013$ is the largest answer choice. So the answer is $\mathbf{(E)}$.

## Alternative Solution

Let $L_k(x)=\log_2\log_2...\log_2(x)$ where there are $k$ $\log_2$'s. $L_k(B)$ is defined iff $L_{k-1}(B) > 0$ iff $L_{k-2}(B) > 1$. Note $\log_2 T(k) = T(k - 1)$, so $L_{k-2}(T(k-2))=1$. Thus, we seek the largest $k$ such that $B > T(k-2)$. Now note that

$T(2009)^{T(2009)^{T(2009)}} > 2^{2^{T(2009)}} = T(2011)$

so $k=2013$ satisfies the inequality. Since it is the largest choice, it is the answer.