Solve x in terms of y.

Question

Question:
y = (e^x + e^-x)/(e^x - e^-x)

Answer:
x= 1/2 ln ((y+1)/(y-1))

I have the answer with me, but I don’t know how to solve the question step by step.

Answer 1

y = ( e^x + e^-x ) / ( e^x - e^-x )

e^-x = 1 / e^x

y = ( e^x + 1 / e^x ) / ( e^x - 1 / e^x )

Replace e^x with u

y = ( u + 1 / u) / ( u - 1 / u)

y = ( u ∙ u / u + 1 / u) / ( u ∙ u / u - 1 / u)

y = ( u^2 / u + 1 / u) / ( u^2 / u - 1 / u)

y = [ ( u^2 + 1 ) / u ] / [ ( u^2 - 1 ) / u ]

y = ( u^2 + 1 ) / ( u^2 - 1 )

Multiply both sides by ( u^2 - 1 )

y ∙ ( u^2 - 1 ) = u^2 + 1

y ∙ u^2 - y = u^2 + 1

Subtract u^2 to both sides

y ∙ u^2 - y - u^2 = u^2 + 1 - u^2

y ∙ u^2 - y - u^2 = 1

Add y to both sides

y ∙ u^2 - y - u^2 + y = 1 + y

y ∙ u^2 - u^2 = y + 1

u^2 ( y - 1 ) = y + 1

Divide both sidess by y - 1

u^2 = ( y + 1 ) / ( y - 1 )

Replace u = e^x in this equation

( e^x )^2 = ( y + 1 ) / ( y - 1 )

Take the natural logarithm of both sides

2 ∙ ln ( e ^ x) = ln [ ( y + 1 ) / ( y - 1 ) ]

since ln ( e ^ x) = x

2 ∙ x = ln [ ( y + 1 ) / ( y - 1 ) ]

Divide both sides by 2

x = ( 1 / 2 ) ln [ ( y + 1 ) / ( y - 1 ) ]

Answer 2

y = (e^x + e^-x)/(e^x - e^-x) = (e^(2x)+1)/(e^(2x)-1)
e^(2x)+1 = e^(2x)*y - y
e^(2x)(y-1) = y+1
e^(2x) = (y+1)/(y-1)
2x = ln (y+1)/(y-1)
x = 1/2 ln (y+1)/(y-1)