Question

y=(2^(x) +1)/ (2^(x) -1)... solve for x.

is x=log_2 ((1+y)/(2-y))?

Answer 1

your equation can become:

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

ln(2^x) = ln[(y+1)/(y-1)]
x = Log_2 [(y+1)/(y-1)]

Answer 2

oh! my bad about the 2-y! it's supposed to be y-1! thank you again!:D

Answer 3

yw;)