In the following argument in which step has a mistake been made:
let logₘn = a
changing the base to 10 gives
a =
a = log₁₀( )
10ᵃ =
however changing the base to 11 instead of 10 gives
11ᵃ =
thus 10ᵃ = 11ᵃ
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Quite a simple mistake has been made in step 2 of this problem:
l o g ₁ ₀ m l o g ₁ ₀ n ≠ log₁₀ m n
A fraction with the logarithms of two values as the numerator and denominator, is not the same as the logarithm of a fraction with those same values as the numerator and denominator.