What Is the Solution to sinh(x) = 1?

  • Thread starter Thread starter kd001
  • Start date Start date
kd001
Messages
42
Reaction score
0

Homework Statement



sinh(x) = 1

What is the value of 'x'?


Homework Equations



sinh(x) = (1/2)(e^x - e^-x)


The Attempt at a Solution



e^x - e^-x = 2

Then what do I do?

Thanks
 
Physics news on Phys.org
Multiply the whole eq by e^{x} then solve the resulting quadtatic eqn in e^{x} and afterwards keep the positive solution and take its logarithm to obtain x.
 
emanuel_hr said:
Multiply the whole eq by e^{x} then solve the resulting quadtatic eqn in e^{x}

That is put, say, u=e^x and solve for u. Then find x from e^x=u.
 
Thanks. That worked.
 
kd001 said:

Homework Statement



sinh(x) = 1

What is the value of 'x'?


Homework Equations



sinh(x) = (1/2)(e^x - e^-x)


The Attempt at a Solution



e^x - e^-x = 2

Then what do I do?

Thanks
Multiply on both sides by e^x to get (e^x)^2- 1= 2e^x and then subtract 2e^x- 1 from both sides: (e^x)^2- 2e^x= 1. Think of that as a quadratic equation in e^x and complete the square: (e^x)^2- 2e^x+ 1= (e^x- 1)^2= 2.

Take the square root of both sides, e^x- 1= \pm\sqrt{2} and add 1 to both sides, e^x= 1\pm\sqrt{2}. 1- \sqrt{2}< 0 so you get the single solution x= ln(1+ \sqrt{2}).

Too slow! Too slow!
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top