4 ms·
> You don't need the ones with a different base. To add on that, we need only one logarithm in the sense that all other then follows. For any base b, we have
by Jenz 6y ago
> You don't need the ones with a different base.
To add on that, we need only one logarithm in the sense that all other then follows. For any base b, we have
log(x; b) = log(x)/log(b)
- 6gvONxR4sf7o 6y agoI think to complete the point, it’s also worth pointing out that for any bases b, c and d: log(x; b) = log(x; c)/log(b; c) = log(x; d)/log(b; d) The second equality is why you can ignore c and d, and just pretend there's one logarithm when you do log(x; b) = log(x)/log(b)