Subset sum vs. Subset product (strong vs. weak NP hardness)
Firstly, using exponentiation to go from SS to SP works (using base 2 rather than base $e$), but blows up the size of the numbers involved. Weak NP-hardness means that if the numbers are small (or more precisely, denoted in unary), the problem is no longer hard. Hence, using exponentiation creates exponentially sized instances of SP even for the easy instances of SS, where the numbers are written in unary.

Secondly, using logarithms to go from SP to SS doesn't work, as logarithms typically generates non-integer values. SS and SP are defined using integer numbers, and logarithms often result in transcendental values, which are hard to represent or do math on.

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Let $A$ be an integer, $A &gt; 0$, then $\log_2 A$ is rational if and only if $A$ is a power of 2, and transcendental otherwise. Firstly, if $\log_2 A = \frac{p}{q}$ for nonzero integers $p$ and $q$, then $A = 2^{\frac{p}{q}}$, $A^q = 2^p$. We therefore have that $A = 2^r$ by prime decomposition. Furthermore $A^{rq} = 2^p$, so given an $A$ we can pick $q=1$ and $p=r$ to prove $\log_2 A$ is rational.

We just need to show that $\log_2 A$ is never transcendental otherwise. This follows from the Gelfond-Schneider theorem, for an equivalent formulation (as can be found on the Wiki page) is "if $\alpha$ and $\gamma$ are nonzero algebraic numbers, and we take any non-zero logarithm of $\alpha$, then $(\log \gamma)/(\log \alpha) = \log_\alpha \gamma$ is either rational or transcendental." It is also easy to verify by taking the converse of the theorem and setting $\alpha^\beta = \gamma$ and hence $\beta = \log_\alpha \gamma$.

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Lastly, consider what happens when we try the dynamic programming algorithm from SS on SP. Because we use products rather than sums, the numbers involved blow up enormously, and the arbitrary precision math required suddenly becomes a factor in the running time. This is why the algorithm cannot solve SP instances quickly even if the numbers are in unary.