Solving ln x for x: ce^{-0.03t}

[cscott]

How does one get from $ln x = -0.03t + c'$ to $x = ce^{-0.03t}$

Why isn't it [itex]x = e^{-0.03t} + c[/tex]? where $c = e^{c'}$

 

[marlon]

How does one get from $ln x = -0.03t + c'$ to $x = ce^{-0.03t}$

Why isn't it [itex]x = e^{-0.03t} + c[/tex]? where $c = e^{c'}$

Well, it cannot be your equation for x, the + should be *

Look at this derivation :

$$ln x = -0.03t + c'$$

$$x = e^{-0.03t + c'}$$

$$x = e^{-0.03t} * e^{c'}$$

They called the $$e^{c'}$$ the constant c.

This can be done since you this constant must be determined using some given initial conditions like at t = 0, x must be 5 or so...

marlon

EDIT : if you want you can just use $$e^{c'}$$ as well, it does not really matter because of the way the constant must be calculated. The result will be the same.

 

[cscott]

Thanks marlon. :)