Question

how do I solve for x from a problem containing hyperbolic cosine? e.g. coshx=6

Answer 1

\[\frac{e^x+e^{-x}}{2}=0\]

Answer 2

\[e^x+e^{-x}=0 => e^{-x}(e^{2x}+1)=0\]

Answer 3

let e^x=y then you end up with a quadratic equation..

Answer 4

maybe convert back to exponentials?\[coshx=6\iff \frac{e^x+e^{-x}}{2}=6\iff e^x+e^{-x}=12\]Multiplying by e^x gives:\[e^{2x}+1=12e^x\iff e^{2x}-12e^x+1=0\]this is a quadratic. use the quadratic formula to get e^x, and youre done.

Answer 5

it is never 0

Answer 6

oh i thought that was a 0 lol

Answer 7

six zero whatever

Answer 8

whatev!

Answer 9

joe has the right plan for you

Answer 10

lol, i feel stupid, I can integrate it, but i cant solve it :)

Answer 11

its not a common question.

Answer 12

yea, I'm kinda in IB math II HL, but i got bored with the curriculum, so im teaching my self some stuff

Answer 13

btw that is the IB equivalent of AP calc BC