$f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$
Solution 1:
Just saying "a branch of logarithm" won't do it. In fact, since the range of $f$ will contain all nonzero complex numbers (see Picard's theorem) you can't choose a particular branch of the logarithm and have $\log f$ be entire.
Hint: $g'(z) = f'(z)/f(z)$.