Question

struggling on hyperbolic functions if any one can help?
i have a question to solve for x 15ch x = -21sh x
i understand that sh x=(e^x -e^-x/2) and ch x =(e^x+e^-x/2) and where ax^2+bx+c=0 but i dont know how to rearage the question? any help greatly appreciated

Answer 1

\[15\cosh x=-21\sinh x\]is the problem?

Answer 2

yes sorry

Answer 3

okay thankyou,

Answer 4

arithmetic error earlier\[-\frac57=\tanh x\]and we have that\[\tanh^{-1}x=\frac12\ln\frac{1+x}{1-x};|x|<1\]which is the case, so we have\[\tanh^{-1}(-\frac57)=\frac12\ln\frac{1-\frac57}{1+\frac57}=\frac12\ln\frac16\]ok that should be it

Answer 5

hmm the solution has gone? how are you getting \[-\left(\begin{matrix}3 \\ 7\end{matrix}\right)\]

Answer 6

that was a typo I fixed above

Answer 7

a very persistent typo, lol

Answer 8

ahh okay so to get -(5/7) you have just dived each by 3? sorry to sound stupid but i would rather understand :)

Answer 9

\[15\cosh x=-21\sinh x\]\[-\frac{15}{21}=\frac{\sinh x}{\cosh x}\]simplifying the fraction and using the identity sinh/cosh=tanh we get\[-\frac57=\tanh x\]from whence follows the above solution

Answer 10

yes, divide top and bottom of the fraction by 3...

Answer 11

ahhh okay sorry i have just look in my book and found \[\tan hx= \left(\begin{matrix}\sin hx \\ \cos hx\end{matrix}\right)\]
is there a way of solving it without inserting logs? (ln)

Answer 12

no, because until we take a logarithm we have\[e^x\]in the problem. There is no way to solve for the real values of x without taking a log that I can see.

Answer 13

by the way, I have no idea why you talk about ax^2+bx+c=0 in your post
as far as I know that has nothing to do with what you're doing

Answer 14

because in previous questions i have done i have used a,b and c as constants, so have applied the quadratic formula

Answer 15

is that wrong to assume?