How Can I Solve This PreCalculus Problem Using Natural Logarithms?

[sh86]

So I was going through some old tests of mine and found this problem in an old precalculus test from long ago. I of course did not get it right and I don't think anyone did. Here's the problem:

Use natural logarithms to solve 2 = [e^x - e^(-x)]/2 for x.

I tried every little trick I know...I have no idea how to get this. Can you believe this was asked on a test in a precalculus course (along with tons and tons of other questions)!?

I don't know if this thread belongs in the homework section or not. It seems like it but this is actually not a homework assignment.

 

[sh86]

OMG! I'd been working on this for so long and then after I posted this message I went to wash my face and a strategy suddenly came to me... Here's what I got:

e^x - e^(-x) = 4
e^(-x) (e^2x-1) = 4
e^2x - 1 = 4e^x
e^2x - 4e^x - 1 = 0
e^x = 2 +/- sqrt(5) by the quadratic formula
x = ln(2 +/- sqrt(5))
but that doesn't work for 2 - sqrt(5) since that's less than 0 so the answer is ln(2+sqrt(5)).

I guess my subconscious is a far better problem solver than I am. (BTW if there's an error anywhere please tell me.) Sorry if this topic is now a waste of space..

 

[sh86]

EDIT: Whoops I guess you deleted your post. Uh...then forget all the stuff I wrote below heh.

 

[courtrigrad]

i made a mistake, you are correct in the way you did it.

 

[matt grime]

Can you believe this was asked on a test in a precalculus course (along with tons and tons of other questions)!?

Yes. This was done in high school "in my day", and that was only a dozen years ago, and still might be in the UK.

 

[mathman]

Let u=e^x (therefore e^-x=1/u). You now have a quadratic equation in u. Solve for u, then x=ln(u).

 

[Jeff Ford]

I thought that looked familiar

[e^x - e^(-x)]/2 = sinh(x)

If you're ok with an approximation, you can solve for x from there.

 

[The Bob]

Yes. This was done in high school "in my day", and that was only a dozen years ago, and still might be in the UK.

We did it at college. This particular exam was only taught to us after we learned hyperbolic functions. The education system in Britain is just getting easier and easier.

The Bob (2004 ©)

 

[mathman]

Use natural logarithms to solve 2 = [e^x - e^(-x)]/2 for x.

Letting u=e^x (see my previous note) allows one to do the above.

 

[Office_Shredder]

We did it at college. This particular exam was only taught to us after we learned hyperbolic functions. The education system in Britain is just getting easier and easier.

The Bob (2004 ©)

Interestingly, the other math majors I'm studying with seem to have covered hyperbolic functions in high school in the UK

 

[Gib Z]

omg obviously australia isn't up to the ranks with any other countries. I am 14 right now and finished my mathamatics and physics courses up to the end of high school, i had to learn hyperbolic functions independantly and still don't know them that good, just the basics ie eqivalent of normal trig functions cept on a hpeybola instead of unit circle...zzz...makes me so sad...

 

[The Bob]

Interestingly, the other math majors I'm studying with seem to have covered hyperbolic functions in high school in the UK

If you have not done Further Maths, in the UK, then, where I am currently studying anyway, you have to learn about hyperbolic functions independently. The maths at school seem very hard at the time but if it was taught in a college sort of way then we would all understand it. The UK is behind with education. I can see this more clearly in Music but that is for a different sub-thread.

The Bob (2004 ©)