It is proved that m Λ Logan = n Λ logam

It is proved that m Λ Logan = n Λ logam

Let x = m ^ Logan
Then take the logarithm with a as the base
loga(x)
=loga[M^logaN]
=logaN*logaM
=logaM*logaN
=loga[N^logaM]
So x = n ^ logam
So m ^ Logan = n ^ logam

If M = n, then logam = Logan?

Logarithmic requirements: base a is greater than zero and not one, true number n is greater than zero
If M = n but both are less than or equal to zero, that representation is wrong

Why is it that if M = n, then logam = Logan is wrong, and vice versa?

Because the true number of logarithm must be a positive number, and m, n do not know whether it is a positive number!