Need help with simple conversion

[kmj764]

(e^x) - (e^-x) / (e^x) + (e^-x) -> (e^2x - 1) / (e^2x + 1)

and

y*e^(2x) + y = e^(2x) - 1 -> e^(2x) = (1+y) / (1-y)

Can you please write down the steps in between?

p.s. I think I am asking too many questions lately. Is it okay? I understand I can't ask every single question in hand but even if only one tenth of them could be answered, it would be indeed a great help, given that I can't afford to hire a personal tutor at the moment. (I graduated from school 6 years ago, so no teacher either) I recently found out about this website because I need to improve my maths skills in a really short period of time (I have 9 months left till I take the university entrance exam in my country, so I study 15 hours a day in average to catch up quick.) I don't know what you guys do for a living, but I really appreciate your intention in helping other people, like answering my stupid questions in no time. Thank you*3(my all time favorite number)^∞

 

[soroban]

Hello, kmj764!

\(\displaystyle \displaystyle \frac{e^x - e^{-x}}{e^x + e^{-x}} \quad\Rightarrow\quad \frac{e^{2x} - 1}{e^{2x} + 1}\)

Muliply by \(\displaystyle \frac{e^x}{e^x}\!:\;\;\dfrac{e^x}{e^x}\cdot \dfrac{e^x - e^{-x}}{e^x + e^{-x}} \;=\; \dfrac{(e^x)(e^x) - (e^x)(e^{-x})}{(e^x)(e^x) + (e^x)(e^{-x})} \;=\;\dfrac{e^{2x} - 1}{e^{2x} + 1} \)

 

[Deleted member 4993]

(e^x) - (e^-x) / (e^x) + (e^-x)

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

Can you please write down the steps in between?

p.s. I think I am asking too many questions lately. Is it okay? I understand I can't ask every single question in hand but even if only one tenth of them could be answered, it would be indeed a great help, given that I can't afford to hire a personal tutor at the moment. (I graduated from school 6 years ago, so no teacher either) I recently found out about this website because I need to improve my maths skills in a really short period of time (I have 9 months left till I take the university entrance exam in my country, so I study 15 hours a day in average to catch up quick.) I don't know what you guys do for a living, but I really appreciate your intention in helping other people, like answering my stupid questions in no time. Thank you*3(my all time favorite number)^∞

This is really simple problem - you need to stare at it with pencil and paper in hand:

\(\displaystyle \dfrac{e^x \ - \ e^{-x}}{e^x \ + \ e^{-x}} \ = \ \dfrac{e^x \ * \ (e^x \ - \ e^{-x})}{e^x \ * \ (e^x \ + \ e^{-x})} \ = \ ???\)

 

[kmj764]

Oops

This is really simple problem - you need to stare at it with pencil and paper in hand:

\(\displaystyle \dfrac{e^x \ - \ e^{-x}}{e^x \ + \ e^{-x}} \ = \ \dfrac{e^x \ * \ (e^x \ - \ e^{-x})}{e^x \ * \ (e^x \ + \ e^{-x})} \ = \ ???\)

You know what, actually I figured that out by myself right before you answered it.

But then I had another question so I edited the question, which happened after your answer.

That is: y*e^(2x) + y = e^(2x) - 1 -> e^(2x) = (1+y) / (1-y)

which is the next step as a matter of fact.

 

[Deleted member 4993]

y*e^(2x) + y = e^(2x) - 1 -> e^(2x) = (1+y) / (1-y)

Put all the terms containg "y" on one side - all the terms without "y" onthe other side of the equation....