Mathematical methods: logarithms question

[StillAnotherDave]

Homework Statement

The question is simply to show that the LHS of the equation equals the RHS and determine the value of A.

Relevant Equations

(e^2x+e^x-1-e^(-x))/(e^x+1)=Asinhx

I know that sinhx = 1/2(e^x-e^-x) and that e^2x-1 = e^x(e^x-e^-x) and similar identities but don't know how to get any further. Any hints at where to go with this would be appreciated.

 

[PeroK]

Homework Statement: The question is simply to show that the LHS of the equation equals the RHS and determine the value of A.
Homework Equations: (e^2x+e^x-1-e^(-x))/(e^x+1)=Asinhx

I know that sinhx = 1/2(e^x-e^-x) and that e^2x-1 = e^x(e^x-e^-x) and similar identities but don't know how to get any further. Any hints at where to go with this would be appreciated.

Hint: try to factorise.

 

[StillAnotherDave]

 

[PeroK]

I guessed that was what it was!

 

[StillAnotherDave]

Yep, I get that the there should be some way to extract a factor but don't know how! Is there a way to take out a factor of e^x + 1 from the numerator for example?

 

[PeroK]

Yep, I get that the there should be some way to extract a factor but don't know how! Is there a way to take out a factor of e^x + 1 from the numerator for example?

Yes there is. What do you have to multiply $e^x + 1$ by to get the numerator?

 

[StillAnotherDave]

Okay, that sounds promising. Let me play around with that and get back to you. Thanks!

 

[WWGD]

It seems too the terms of the numerator form a G.P. Maybe that will help.

 

[DEvens]

It's curious that the title of the thread is "logarithms question" but there are no logs in the question.

 

[StillAnotherDave]

It's curious that the title of the thread is "logarithms question" but there are no logs in the question.

Very true!

 

[StillAnotherDave]

Thanks all for the help. Once I realized (the obvious) that "1" can be expressed in terms of e things resolved themselves.

 

[Mark44]

It seems too the terms of the numerator form a G.P. Maybe that will help.

No, that doesn't help. The terms would form a geometric progression if all of them were the same sign, but that's not the case here.

 

[Mark44]

Thanks all for the help. Once I realized (the obvious) that "1" can be expressed in terms of e things resolved themselves.

It's simpler to just factor things. The numerator is $e^{2x} + e^x - 1 - e^{-x} = e^x(e^x + 1) - e^{-x}(e^x + 1) = (e^x - e^{-x})(e^x + 1)$

 

[StillAnotherDave]

It's simpler to just factor things. The numerator is $e^{2x} + e^x - 1 - e^{-x} = e^x(e^x + 1) - e^{-x}(e^x + 1) = (e^x - e^{-x})(e^x + 1)$

Yep. That's what I meant. Initially I didn't see that $e^{-x}e^{x}$ is equivalent to 1.

 

[WWGD]

No, that doesn't help. The terms would form a geometric progression if all of them were the same sign, but that's not the case here.

Or if the signs were alternating, but that's not the case either. Will have my espresso _ before_ PF'ing from now on :).