# Proof of equation $\ln y=-kt+c$ is same as $y=c*e^{-kt}$

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.

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

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 $y =c e^{-kt}$

• It was part of the ODE example, yes. Thank you. Commented Jul 5, 2016 at 2:14

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.

• 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. Commented Jul 5, 2016 at 2:18
• @JekDenys no problem, I remember it confusing me when I first saw it too Commented Jul 5, 2016 at 2:19