# SICP Exercise 1.26

Let argument `exp`

be `n`

. The number of steps for iterative
process can be expressed by recurrency \(T(n)=T(n/2)+O(1)\), with
solution \(O(\log{}n)\).

Louis replaced fast iterative process with recursive one, that
splits in two branches for even `n`

. This corresponds to different
recurrency \(T(n)=2 T(n/2)+O(1)\), with solution \(O(n)\).