How do I solve for x? e^x - 6e^-x = 1.

Question

I tried to factor it out and I did: e^x (1 - 6e^-2x) = 1 however I could not get an answer.

Answer 1

e^x - 6ex^-1 = 1

e^x - 6/e^x = 1

let y = e^x

y - 6/y = 1

multiply by y

y^2 - 6 = y

y^2 -y -6 = 0

factor

(y-3)(y+2) = 0

y = 3 = e^x

ln3 = x

and

y =-2 = e^x

ln-2 = x but logs can't be of negative numbers

so just one solution x= ln3

Answer 2

e^x - 6 / e^x = 1

Multiply by e^x: (e^x)² - 6 = e^x

(e^x)² - e^x - 6 = 0

(e^x - 3)(e^x + 2) = 0

e^x = 3 or e^x = -2

Since e^x is always positive, the equation e^x = -2 has no solution.

So, e^x = 3. x = ln3