I just had this problem come up at work, as part of a simulation where I had to solve the equation mentioned above (where $m$ and $k$ are constants). I googled solving exponential equations and I got so far as realizing that I need to log both sides of the equation resulting in:
$$2\ln x = -mx + \ln k$$
The above form of the equation seems more intractable than the first and am at a loss regarding how to proceed. Can someone please give me a hint as to the way forward?
Answer
When you have exponential and linear or quadratic function in same equation, you must use Lambert-$\operatorname{W}$ function which is defined as inverse function of $f(x)=e^xx$.
$$2\ln x=-mx+\ln k$$
$$x^2=e^{-mx}k$$
$$e^{-mx}k=x^2$$
$$e^{-mx}x^{-2}=\dfrac1k$$
$$e^{\dfrac{mx}2}x=\sqrt k$$
$$e^{\dfrac{mx}2}\dfrac{mx}2=\dfrac{m\sqrt k}2$$
$$\dfrac{mx}2=\operatorname{W}_k\left(\dfrac{m\sqrt k}2\right),k\in\mathbb{Z}$$
$$x=\dfrac{2\operatorname{W}_k\left(\dfrac{m\sqrt k}2\right)}{m}$$
No comments:
Post a Comment