If logam = Logan, then why is m = n right

If logam = Logan, then why is m = n right

Because y = logax has monotonicity in the domain
It shows that X and y are one-to-one correspondence
So the same y, only the same X
So if logam = Logan, then M = n

m∧logaN=n∧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 2loga (m-2n) = logam + Logan, calculate M / n

∵2loga(M-2N)=logaM+logaN,
And 2loga (m-2n) = loga (m-2n) ^ 2
logaM+logaN=logaMN
∴(M-2N)^2=MN
∴M^2-5MN+4N^2=0
That is, (m-n) (m-4n) = 0
M / N = 1 or M / N = 4
And ∵ m-2n ＞ 0,
‖ M / N = 1, rounding off
To sum up, M / N = 4