2ln(x-1)=ln(x^2-5) solve for x

Question

Could you help me with solving this solution for x please?

Answer 1

2ln(x-1)=ln(x^2-5) solve for x

ln(x - 1)^2 - ln( x^2 - 5) = 0

ln((x -1)^2)/(x^2 - 5) = 0

e^0 = ( x - 1)^2/(x^2 - 5 )

1 = ( x - 1)^2/(x^2 - 5 )

x^2 - 5 = ( x - 1)^2

x^2 - 5 = x^2 - 2x + 1

x^2 - 5 - x^2 + 2x  - 1 = 0

2x - 6 = 0

2x = 6

x = 3

Answer 2

proof that (3)/(ln(4))+(4)/(ln(5))≥(2)/(ln(3))+(5)/(ln(6))