The discussion focuses on the mathematical derivation of the equation x = ce^{-0.03t} from ln x = -0.03t + c'. The transformation involves recognizing that e^{c'} is a constant, denoted as c, which can be determined using initial conditions. The incorrect assumption that x could be expressed as e^{-0.03t} + c is clarified, emphasizing the need for multiplication rather than addition in the exponential form. The derivation is crucial for solving differential equations involving exponential decay.
PREREQUISITESStudents of mathematics, educators teaching calculus, and professionals working with differential equations or mathematical modeling will benefit from this discussion.
[itex] <br /> Well, it cannot be your equation for x, the + should be *<br /> <br /> Look at this derivation :<br /> <br /> [tex]ln x = -0.03t + c'[/tex]<br /> <br /> [tex]x = e^{-0.03t + c'}[/tex]<br /> <br /> [tex]x = e^{-0.03t} * e^{c'}[/tex]<br /> <br /> They called the [tex]e^{c'}[/tex] 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 [tex]e^{c'}[/tex] as well, it does not really matter because of the way the constant must be calculated. The result will be the same.[/itex]cscott said:How does one get from [itex]ln x = -0.03t + c'[/itex] to [itex]x = ce^{-0.03t}[/itex]
Why isn't it [itex]x = e^{-0.03t} + c[/tex]? where [itex]c = e^{c'}[/itex][/itex]