0
$\begingroup$

I'm trying to prove that one is the same as the other : $$\ln y = -kt+c$$ $$y=ce^{-kt}$$

Where c is undefined and k is defined constant. I got as far as: $$y=e^{-kt+c}$$ So by what rule would c be multiplied by e? Could someone explain please? Thank you.

$\endgroup$
4
  • 1
    $\begingroup$ $$e^a \cdot e^b = e^{a+b}$$ Thus, $e^{-kt+c} = e^{-kt} \cdot e^c$ Therefore your statement is only true if $e^c = c$ $\endgroup$
    – StopReadingThisUsername
    Commented Jul 5, 2016 at 2:07
  • 1
    $\begingroup$ This is what I've been looking for. Thank you. $\endgroup$
    – Jek Denys
    Commented Jul 5, 2016 at 2:10
  • $\begingroup$ Please use $ equation $, even in titles. $\endgroup$
    – user64742
    Commented Jul 5, 2016 at 2:10
  • $\begingroup$ They are not quite equivalent. For example one cannot quite get $y=-5e^{-kt}$ from any (real) $\ln y=-kt+c$. $\endgroup$
    – André Nicolas
    Commented Jul 5, 2016 at 2:10

2 Answers 2

Reset to default
1
$\begingroup$

They are not the same, but usually, when one is solving a differential equation (I can't imagine another instance where these two would be "equivalent" in some sense), one writes:

$$\ln y = -kt + c \\ y = e^{-kt + c} \\ y = e^c e^{-kt}$$

Now $e^c$ is just a constant, so we may have given $c$ the name $a$ and then we would let $c = e^a$, so that $y =c e^{-kt}$

$\endgroup$
1
  • $\begingroup$ It was part of the ODE example, yes. Thank you. $\endgroup$
    – Jek Denys
    Commented Jul 5, 2016 at 2:14
0
$\begingroup$

That's because they aren't! Usually, in the context of differential equations, the $A$ in $$ y=Ae^{-kt} $$ Really is a more concise stand in for some exponentiated constant, in your case $e^c$, i.e. $$ \ln(y)=-kt+c\Rightarrow y=e^{-kt+c}=e^ce^{-kt}\Rightarrow y=Ae^{-kt} $$

You can view the $A$ as absorbing constants you are adding on.

$\endgroup$
2
  • $\begingroup$ Thank you. It was really the operation with the $e^{a+b} = e^{a}e^{b}$ that I was looking for. Completely blanked out on it. $\endgroup$
    – Jek Denys
    Commented Jul 5, 2016 at 2:18
  • $\begingroup$ @JekDenys no problem, I remember it confusing me when I first saw it too $\endgroup$
    – operatorerror
    Commented Jul 5, 2016 at 2:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .