+26387 How do you solve this equation?
\(\dfrac{1}{e^x - e^{-x}}= 2 \)
\(\begin{array}{rcl} \dfrac{1}{e^x - e^{-x}} &=& 2 \\ e^x - e^{-x} &=& \dfrac{1}{2} \qquad | \qquad \dfrac{e^x - e^{-x}}{2} = \sinh{(x)}\\ 2\sinh{(x)} &=& \dfrac{1}{2} \\ \sinh{(x)} &=& \dfrac{1}{4}\\ x &=& \text{arsinh}{ \left(\dfrac{1}{4} \right)} = \text{sinh}^{-1}{ \left(\dfrac{1}{4} \right)}\\ \mathbf{x} & \mathbf{=} & \mathbf{ 0.2474664615472692 } \end{array}\)
+118667 Come on Namadesto,
We have answered 2 of these questions for you in the last 12 hours.
How about you giving us some of your thoughts on how you might be able to solve it ?
We can talk those through with you if you like :)
Hint: let y=e^x
You can simplify before or after you do the substitution. :)
+129845 (1/(ex - e-x)) = 2
Notice that we can invert both sides and get :
(ex - e-x) = 1/2 multiply through by 2ex
2e2x - 2 = ex subtract ex from both sides and rearrange
2e2x - ex - 2 = 0 this will not factor......let ex = y and we have
2y^2 - y - 2 = 0 and the solutions for this are....y ≈ -.78078 and y = 1.2808
Therefore, ex = -.78078 (impossible)....... or ex = 1.2808
So
ex = 1.2808 which means that x = ln 1.2808 = about .2475
+26387 How do you solve this equation?
\(\dfrac{1}{e^x - e^{-x}}= 2 \)
\(\begin{array}{rcl} \dfrac{1}{e^x - e^{-x}} &=& 2 \\ e^x - e^{-x} &=& \dfrac{1}{2} \qquad | \qquad \dfrac{e^x - e^{-x}}{2} = \sinh{(x)}\\ 2\sinh{(x)} &=& \dfrac{1}{2} \\ \sinh{(x)} &=& \dfrac{1}{4}\\ x &=& \text{arsinh}{ \left(\dfrac{1}{4} \right)} = \text{sinh}^{-1}{ \left(\dfrac{1}{4} \right)}\\ \mathbf{x} & \mathbf{=} & \mathbf{ 0.2474664615472692 } \end{array}\)
