Understanding Laws of Logs: How to Derive Chemistry Kinetics Equation

[mycotheology]

I'm reading about how the chemistry kinetics equations are derived and here's 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.

 

[n_kelthuzad]

I'm reading about how the chemistry kinetics equations are derived and here's 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 let's say :
ln a = ln b - c
e^(ln a) = e^(ln b - c)
u got your exponentialtion wrong.

 

[scurty]

I'm reading about how the chemistry kinetics equations are derived and here's 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.

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