[salmon12345]

Find the real solutions of 3*sinh(x/2) -cosh(x/2)=1

I subbed in sinh(x/2)=((exp(x/2) -exp(-x/2))/2) and cosh(x/2)=((exp(x/2) +exp(-x/2))/2)

got down to exp(x/2)-2*exp(-x/2)=1

then ln(exp(x/2)) - 2ln(exp(-x/2))=ln(1) (***)

getting 3*x/2=0 therefore x=0 which doesn't work when subbed back in to original equation.

I'm guessing something is wrong with (***) line?

[jack.hadamard]

(Original post by salmon12345)

got down to exp(x/2)-2*exp(-x/2)=1

then ln(exp(x/2)) - 2ln(exp(-x/2))=ln(1) (***)

You cannot do this. Think of quadratic equations at this stage.

[Intriguing Alias]

(Original post by salmon12345)
Find the real solutions of 3*sinh(x/2) -cosh(x/2)=1

I subbed in sinh(x/2)=((exp(x/2) -exp(-x/2))/2) and cosh(x/2)=((exp(x/2) +exp(-x/2))/2)

got down to exp(x/2)-2*exp(-x/2)=1

then ln(exp(x/2)) - 2ln(exp(-x/2))=ln(1) (***)

getting 3*x/2=0 therefore x=0 which doesn't work when subbed back in to original equation.

I'm guessing something is wrong with (***) line?

If you wanted to natural log you have to natural log the whole side, not term by term. i.e. if we have x + y = 1 then ln(x+y) = ln1, but this does NOT mean lnx + lny = 1.

Anyway; don't natural log. After the bolded, multiply all terms by e^(x/2) and see where you can go from there.

[Hasufel]

you can boil the equation down to p^2-p-2=0, where p=Exp[x/2]

factorise this - and you might spot a VERY FAMOUS Identity....(which will give ONE result, but not a real one)

Edit: the other factor should give you the real result - if you take (natural) logs of both sides of it...

Edit (these will be the principal solutions, like "principal arguments")

[salmon12345]

Thanks, I think I've got it.