Question

show all working for -1/2e^(x/2) + 1/2e^(-x/2) = -1 solve for x
please help me! :'|

Answer 1

I have a CAS and know that the answer is x = 2ln(1+ 2^1/2) but i have to show all working out :(

Answer 2

let u = x/2 and rearrange the order, yours is
\[\frac{1}{2e^{-u}}-\frac{1}{2e^u}=-1\\\frac{e^u}{2}-\frac{e^{-u}}{2}=-1\\\frac{e^u-e^{-u}}{2}=-1\\sinh u =-1 \\u =sinh^{-1}(-1) \]but I really not know whether you know it or not, still replace u =x/2 to solve for x

Answer 3

ok, I have other way to come up with your problem
from \[\frac{e^u-e^{-u}}{2}=-1\\e^u -e^{-u}=-2\\e^u-\frac{1}{e^u}=-2\]make the same denominator and time both sides by \(e^u\), you have \[(e^u)^2-1=-2e^u\]
\[(e^u)^2 +2e^u -1=0\]
now let e^u =t ,so \[t^2 +2t -1=0\]solve for t, then solve for u then solve for x.
No choice for you when you don't know how to solve sinh / cosh.

Answer 4

t= -1\(\pm\sqrt{2}\)
first case, t = -1 +\(\sqrt{2}\), \(e^u=-1+\sqrt{2}\rightarrow u =ln(-1+\sqrt{2})\)
so, x = 2 ln (-1+\(\sqrt{2})\)do the same with other value of t. good luck

Answer 5

thank you! you're brilliant!