Answer: E

Small Hint:

Repeatedly applying $\log_2$ peels one level off a tower: $\log_2 T(n + 1) = T(n)$

Big Hint:

Track how many logs $B = (T(2009))^A$ can survive before the value drops below $1$

Solution:

Since $\log_2 T(n + 1) = T(n),$ each application of $\log_2$ strips one $2$ off the top of a tower of twos.

Write $T_j=T(j),$ and let $L_j$ be the result of applying $\log_2$ to $B$ exactly $j$ times. The first two results are $$\begin{aligned} L_1 &= A\,T_{2008}, \\ L_2 &= T_{2009}T_{2008}+T_{2007}. \end{aligned}$$

For the lower bound, $$\begin{aligned} L_3 &\gt \log_2(T_{2009}T_{2008}) \\ &= T_{2008}+T_{2007} \\ &\gt T_{2008}. \end{aligned}$$ Repeatedly taking logarithms gives $L_{k+3} \gt T_{2008-k}$ for $0\le k\le 2007.$ In particular, $L_{2010}\gt2,$ so $L_{2011}\gt1$ and $L_{2012}\gt0.$ Thus $L_{2013}$ is defined.

For the upper bound, $T_{2007}\lt T_{2008}T_{2009},$ so $$\begin{aligned} L_3 &\lt 1+T_{2007}+T_{2008} \\ &\lt 2T_{2008}, \\ L_4 &\lt 1+T_{2007} \lt T_{2008}. \end{aligned}$$ Repeating the last comparison gives $L_{k+4}\lt T_{2008-k}$ for $0\le k\le2007.$ Hence $L_{2011}\lt2.$ Together with the lower bound, this yields $0\lt L_{2012}\lt1,$ so $L_{2013}\lt0.$ Therefore $L_{2014}$ is undefined, and the largest possible $k$ is $2013.$

Thus, the correct answer is E.