How do I solve for K in ln(1-4k) = -6k?

CursedAntagonis
Messages
23
Reaction score
0

Homework Statement


Trying to solve for K:

ln(1-4k)=-6k


Homework Equations





The Attempt at a Solution


I know that need to take the e of both side, to get 1-4k=e^-6k, but I cannot remember any properties of e to allow me to remove the k from the exponent. Any tips is appreciated.
 
Physics news on Phys.org
CursedAntagonis said:

Homework Statement


Trying to solve for K:

ln(1-4k)=-6k


Homework Equations





The Attempt at a Solution


I know that need to take the e of both side, to get 1-4k=e^-6k, but I cannot remember any properties of e to allow me to remove the k from the exponent. Any tips is appreciated.

Clearly, k=0 is a solution.
to find the another solution you need to use an advanced topic:
http://mathworld.wolfram.com/LambertW-Function.html
 
You have to solve it numerically. You can't solve for k algebraically.

You can visualize the solutions by plotting the functions y=1-4k and y=e^-6k and seeing where they intersect.
 
CursedAntagonis said:
the answer is actually 0.1457 m/s.

m/s?? :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top