Laws of logs question


by mycotheology
Tags: laws, logs
mycotheology
mycotheology is offline
#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.
Phys.Org News Partner Mathematics news on Phys.org
Math modeling handbook now available
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
n_kelthuzad
n_kelthuzad is offline
#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
scurty is offline
#3
Apr23-12, 09:30 PM
P: 380
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*}
##


Register to reply

Related Discussions
Laws of logs Precalculus Mathematics Homework 4
Algebra: Laws of Logs General Math 9
calculus with logs and natural logs Calculus & Beyond Homework 4
logs question Introductory Physics Homework 10
Question on logs General Math 2