Continuous logarithms in cryptography
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.