Question

How did (e^x-e^-x)/(e^x+e^-x) become (e^2x-1)/ (e^2x+1) ??

Answer 1

that is done by multiplying the numerator and denominator by \(e^x\)

Answer 2

\[\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}*\frac{e^x}{e^x}\]\[=\frac{e^x(e^x-e^{-x})}{e^x(e^x+e^{-x})}=\frac{e^{2x}-1}{e^{2x}+1}\]

Answer 3

Now I get it!! Thanks! Last question, why do I need to multiply it to e^x

Answer 4

I assume this is related to your last question on \(\tanh{x}=\frac{1}{2}\)?

Answer 5

Yes

Answer 6

ok, so you know that:\[\tanh{x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\]
correct?

Answer 7

Yes

Answer 8

we are therefore trying to solve:\[\tanh{x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{1}{2}\]

Answer 9

one way of doing that is to first simplify the left-hand-side

Answer 10

so that you are left with a fraction involving just \(e^{2x}\)

Answer 11

you saw how that was done, so we then end up with:\[\frac{e^{2x}-1}{e^{2x}+1}=\frac{1}{2}\]now you can just cross multiply and solve for \(e^{2x}\)

Answer 12

does that make sense?

Answer 13

Yes! Thanks a lot!!

Answer 14

yw :)