solving log equations for x?

Question

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

x = ..?

i am totally lost on this problem...thanks for your help!

Answer 1

If u = e^x, then 1/u = 1/e^x = e^(-x)

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

Multiply both sides by (u + 1/u)

u - 1/u = -t (u + 1/u) = -ut - t/u

Multiply both sides by u

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

Re-substitute u=e^x:

e^x = +/-sqrt[(1-t)/(1+t)]

Take ln of both sides:

x = ln[sqrt[(1-t)/(1+t)]

Sorry I screwed that up the first time. I need sleep. Thanks for getting it right, Sidarth.

Answer 2

put e^x=y
therefore
(y-1/y)/y+1/y)=-t
y^2-1=-t(y^2) -t
y^2=(1-t)/1+t=k
e^2x=k
2x=lnk
x=.5lnk
thanks

Answer 3

(e^x-(1/e^x)/(e^x+(1/e^x)=-t
(e^2x-1)/(e^2x+1)=-t
e^2x=te^2x=-t+1
e^2x=(-t+1)/(t+1)
use logarithms

Answer 4

(e^x - e^-x) / (e^-x + e^x) = -t
tanh x=-t
x=arctanh -t