3 ms·
> How can it be faster considering that they are equivalent? Computing `ln(e^x)` and `x` are equivalent, but the latter is faster.
by JadeNB 7y ago
> How can it be faster considering that they are equivalent?
Computing `ln(e^x)` and `x` are equivalent, but the latter is faster.
- Koshkin 7y agoWell, to be fair, these are equal, not just equivalent.