# Laws of logs question

I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
get turned into this:
$[A] = [A]_{0}*e^{-kt}$?
When I try to derive it, I first get this:
$ln[A] - ln[A]_{0} = -kt$.
Then I isolate ln[A] and get:
$ln[A] = ln[A]_{0} - kt$
then I reverse the ln on both sides of the equation and get:
$[A] = [A]_{0} - e^{-kt}$.
I don't understand how the two terms end up multiplied rather than subtracted.

I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
get turned into this:
$[A] = [A]_{0}*e^{-kt}$?
When I try to derive it, I first get this:
$ln[A] - ln[A]_{0} = -kt$.
Then I isolate ln[A] and get:
$ln[A] = ln[A]_{0} - kt$
then I reverse the ln on both sides of the equation and get:
$[A] = [A]_{0} - e^{-kt}$.

I don't understand how the two terms end up multiplied rather than subtracted.
The red part is the wrong part.
By reverse you use e as a index for exponentiation, so lets say :
ln a = ln b - c
e^(ln a) = e^(ln b - c)

I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-6.gif
get turned into this:
$[A] = [A]_{0}*e^{-kt}$?
When I try to derive it, I first get this:
$ln[A] - ln[A]_{0} = -kt$.
Then I isolate ln[A] and get:
$ln[A] = ln[A]_{0} - kt$
then I reverse the ln on both sides of the equation and get:
$[A] = [A]_{0} - e^{-kt}$.
I don't understand how the two terms end up multiplied rather than subtracted.
The easiest method to see this is to apply the exponential to each side.

##
\begin{eqnarray*}
\displaystyle ln\frac{[A]}{[A]_0} &=& -kt\\
\displaystyle e^{ln\frac{[A]}{[A]_0}}&=& e^{-kt}\\
\displaystyle \frac{[A]}{[A]_0} &=& e^{-kt}\\
\displaystyle [A] &=& [A]_0 e^{-kt}\\
\end{eqnarray*}
##