Simple log/exponential equation.

[Willowz]

HI, I'm having problems dealing with this simple equation. I just need to know what steps need to be taken to solve it.

e^2x - e^-x = 1

Answer.

x = ln(1+$$\sqrt{5}$$/2)

Thanks.

 

[Char. Limit]

Multiply both sides by e^(x) and then you'll get a cubic equation in e^(x) to solve.

 

[Dick]

Sure, but cubics are hard to solve. And the solution of the cubic isn't ln(1+sqrt(5)/2). I think the problem is supposed to read e^(2x)-e^(x)=1.

 

[Char. Limit]

Sure, but cubics are hard to solve. And the solution of the cubic isn't ln(1+sqrt(5)/2). I think the problem is supposed to read e^(2x)-e^(x)=1.

Good point. That's a quadratic in e^(x), and that's much easier to solve...

 

[Willowz]

Sorry there. The equation reads. e^x-e^-x=1.

 

[Char. Limit]

Multiply both sides by e^(x) and then substitute u=e^(x). You'll get a quadratic in u, solve for u, then substitute e^x back in.

 

[Willowz]

Thanks!. I got u^2-u-1=0 and the rest was cake.

 

[HallsofIvy]

Or note that $e^x- e^{-x}= 2 cosh(x)= 1$