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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top