# Understanding exponential computation by digit recurrence

I've met in a book the following algorithm that computes the exponential:

Input: $t, n$ ($n$ is the number of steps)

Output: $E_n$

$\begin{array}{l} \mbox{define$t_0 = 0$;$E_0 = 1$} \\ \mbox{build two sequences$t_i$and$E_i$as follows} \\ t_{n+1} = t_n + ln(1 + d_n 2^{-n}) \\ E_{n+1} = E_n(1 + d_n 2^{-n}) \\ d_n = \begin{cases} 1 & \mbox{if$t_n + ln(1 + 2^{-n}) \leq t$} \\ 0 & \mbox{otherwise} \end{cases} \end{array}$

It is taken from this book. My question is not theoretical (it resembles the restoring division, or square root, or reciprocal) but implementative. To me it looks like that such algorithm requires a LUT that stores the values of $ln(1 + 2^{-n})$ while the others algorithms i mentioned don't... Does the algorithm actually requires a LUT to be implemented? or could be somehow exploited the fact that $ln(1 + 2^{-n}) \approx 2^{-n}$?