I don't know how hard they would be to come up with ex nihilo, but they're fairly straightforward to prove if you know that logb(x) = log(x)/log(b) and log(x^y)

I don't know how hard they would be to come up with ex nihilo, but they're fairly straightforward to prove if you know that logb(x) = log(x)/log(b) and log(x^y)=y * log(x). For logy√z(z) = y logy√z(z) = log(z)/log(z^1/y) = log(z)/(1/y × log(z)) = y * log(z)/log(z) = y For logx(z)√z = x logx(z)√z = z^(1/(logx(z))) log(LHS) = log(z^(1/logx(z)) = log(z)/logx(z) = log(z)/(log(z)/log(x)) = log(x) = log(RHS) Therefore LHS = RHS