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 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 x = e^{-0.03t} + c[/tex]? where c = e^{c'}

<br /> <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&amp;#039;<br /> <br /> x = e^{-0.03t + c&amp;#039;}<br /> <br /> x = e^{-0.03t} * e^{c&amp;#039;}<br /> <br /> They called the e^{c&amp;#039;} 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&amp;#039;} 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. :)