> 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)
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)
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)