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Laws of logs question

by mycotheology
Tags: laws, logs
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mycotheology
#1
Apr23-12, 08:25 PM
P: 90
I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:

get turned into this:
[itex][A] = [A]_{0}*e^{-kt}[/itex]?
When I try to derive it, I first get this:
[itex]ln[A] - ln[A]_{0} = -kt[/itex].
Then I isolate ln[A] and get:
[itex]ln[A] = ln[A]_{0} - kt[/itex]
then I reverse the ln on both sides of the equation and get:
[itex][A] = [A]_{0} - e^{-kt}[/itex].
I don't understand how the two terms end up multiplied rather than subtracted.
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n_kelthuzad
#2
Apr23-12, 08:32 PM
P: 26
Quote Quote by mycotheology View Post
I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:

get turned into this:
[itex][A] = [A]_{0}*e^{-kt}[/itex]?
When I try to derive it, I first get this:
[itex]ln[A] - ln[A]_{0} = -kt[/itex].
Then I isolate ln[A] and get:
[itex]ln[A] = ln[A]_{0} - kt[/itex]
then I reverse the ln on both sides of the equation and get:
[itex][A] = [A]_{0} - e^{-kt}[/itex].

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)
u got your exponentialtion wrong.
scurty
#3
Apr23-12, 09:30 PM
P: 392
Quote Quote by mycotheology View Post
I'm reading about how the chemistry kinetics equations are derived and heres something I don't get. How does this:

get turned into this:
[itex][A] = [A]_{0}*e^{-kt}[/itex]?
When I try to derive it, I first get this:
[itex]ln[A] - ln[A]_{0} = -kt[/itex].
Then I isolate ln[A] and get:
[itex]ln[A] = ln[A]_{0} - kt[/itex]
then I reverse the ln on both sides of the equation and get:
[itex][A] = [A]_{0} - e^{-kt}[/itex].
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*}
##


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