Continuous logarithms in cryptography Cryptographic procedures seem to almost exclusively use discrete logarithms rather than continuous logarithms. Hence, I assume there are good and sound reasons for this. 

In essence, answers provided here: 
Why is NON DISCRETE logarithm problem not hard as the DISCRETE logarithm problem (so computationally hard)? 
 state that continuous logarithms, thus, non-discrete logarithms can be computationally fast, while for discrete logarithms no such algorithm is known.

Furthermore, continuous logs can pose the problem of mathematical precision in terms of rounding errors. 

For any non-discrete logarithm problem such as:

$$v = \log_{b}e$$

solving for either an unknown base $b$ or an unknown exponent $e$ is trivial when $v$ is known.

So my question is, not considering rounding issues, for now, what about solving for $v$ in the equation above, however when both the base and exponent are unknown, with the base $b$ being prime and $e$ being a random integer greater than 1.
By uniqueness of factorisation, this problem is well defined. If $e$ is uniformly distributed as you suggest it must be uniformly distributed in some finite range, say in $(1,2^N)$. Rewrite the equation as 
$$
v \log_2 b = \log_2 e.
$$
Note that the random variable $Z=\log_2 e$ is distributed in $(0,N]$ with a highly nonuniform distribution exponentially biased towards the high end, namely
$$
\mathbb{P}(Z\in (s-1,s)) \propto 2^{N-s+1}
$$
while the primes from which $b$ is drawn are distributed (by the prime number theorem) result in the distribution
$$
\mathbb{P}(W \in (f-1,f)) \propto 
\frac{\log(f-1)}{\log f} 2^{N-f+1}
$$
where $W=\log_2 b.$ Note that $s,f$ are reals not integers.

One can set up a faster than brute force search algorithm given $v$ by first searching for $(f,s)$ pairs with $s=f/v$ starting with $f=N$ and decreasing $f$ while testing the nearest integer to $2^{f/v}$ for primality.

Rewriting the original equation $b^v=e$ as the approximation
$$
\left| \frac{1}{b}-\frac{e}{b^{v+1}}\right|&lt;\frac{1}{2 b^{v+1}}, 
$$
a $v+1$ step continued fraction approximation will be enough to run, prior to primality testing the nearest integer to the approximation.