I 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 eq

I 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)