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 $c = e^{c'}$[/itex]

 

[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 $c = e^{c'}$[/itex]

[itex] <br /> Well, it cannot be your equation for x, the + should be *<br /> <br /> Look at this derivation :<br /> <br /> $$ln x = -0.03t + c'$$<br /> <br /> $$x = e^{-0.03t + c'}$$<br /> <br /> $$x = e^{-0.03t} * e^{c'}$$<br /> <br /> They called the $$e^{c'}$$ the constant c.<br /> <br /> 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...<br /> <br /> marlon<br /> <br /> 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.[/itex]

 

[cscott]

Thanks marlon. :)