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

Question

asnaseer · Answer

that is done by multiplying the numerator and denominator by $e^x$

asnaseer · Answer

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

anonymous · Answer

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

asnaseer · Answer

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

anonymous · Answer

Yes

asnaseer · Answer

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

anonymous · Answer

Yes

asnaseer · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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 $e^{2x}$

asnaseer · Answer

does that make sense?

anonymous · Answer

Yes! Thanks a lot!!

asnaseer · Answer

yw :)