Question

isolate x

Answer 1

yeh thats easy , hyperbolic cos

Answer 2

multiply through by 2e^x

Answer 3

by using logarithms and exponential properties. Thanks

Answer 4

you get a quadratic eqn

Answer 5

2ye^x = (e^x)^2 +1

(e^x)^2 -2y(e^x) +1 =0

Answer 6

u=e^x

Answer 7

e^x = t
(t^2 + 1)/2t=y

Answer 8

u^2 -2yu +1 =0\[u= \frac{ 2y +- \sqrt{4y^2 -4 } } {2} \]

Answer 9

\[u= \frac{ 2y +- 2 \sqrt{ y^2-1} } {2} \]

Answer 10

why to multiply be 2e^x?

Answer 11

\[e^x = \frac{ 2y +- 2\sqrt{y^2 -1}} {2}\]

Answer 12

but e^x >0 for all x , so take the + case

Answer 13

\[e^x = \frac{ 2y + 2\sqrt{y^2-1} } {2} = y + \sqrt{y^2 -1} \]

Answer 14

\[x= \ln ( y + \sqrt{y^2-1} )\]

Answer 15

how do you get that? I dont get it
e^x = t
(t^2 + 1)/2t=y

Answer 16

:|

Answer 17

replace e^x by t

Answer 18

u get

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

Answer 19

let me figure it out :)

Answer 20

got it. Thanks guys