Computing the number of bits of a large power of integer
[edit] As suggested, I edit my answer to give more details.

The answer to my second question is no:

Proposition. Computing $\log(x)$ up to precision $k$ is at least as hard as computing the bit-size of $x^{2^k}$.

Proof. Let $|y|$ denote the bit-size of an integer $y$. First notice that for a nonnegative integer $y$, the bit-size of $y$ is $1+\lfloor \log y\rfloor$. 

Thus, $\left|x^{2^k}\right| = 1+\lfloor 2^k\log x\rfloor$. Now $2^k\log(x)$  is $\log(x)$ shifted $k$ positions to the left. Thus one can compute $\log(x)$ to precision $k$ by simply subtracting $1$ to the bit-size of $x^{2^k}$ and shifting the result $k$ positions to the right.