What is the algebraic solution to: (e^x + e^-x) / (e^x - e^-x) = 3?

Question

for my algebra and trig class

Answer 1

(e^x + e^-x) / (e^x - e^-x) = 3
(e^x + e^-x) = 3e^x - 3e^-x
-2e^x = -4e^-x
e^x = 2e^-x
e^(2x) = 2
2x = ln2
x = ln(2)/2

Answer 2

If you call e ^x = z
you get (z+1/z)/(z-1/z)= 3 so z^2+1 =3z^2-3 and
2z^2 = 4 z=+-sqrt(2) but as z must be >0
e^x = sqrt(2) so x =1/2 ln(2)

Answer 3

Firstly, note that the left hand side is just coth(x) or the hyperbolic cotangent function. so the expression reduces to:
coth(x) = 3
A quick google search reveals that coth^(-1)(x) = 1/2 * ln((x + 1)/(x-1))
so x = coth^(-1)(3) = 1/2 * ln(4/2) = ln(2)/2 as required

Answer 4

Multiply each side by the denominator

e^x + e^-x = 3e^x - 3e^-x

4e^-x = 2e^x

2e^-x = e^x

Take the ln of each side

ln(2e^-x) = x

Use log properties

ln(2) + ln(e^-x) = x

ln(2) - x = x

ln(2) = 2x

0.5*ln(2) = x