# Laws of logs question

by mycotheology
Tags: laws, logs
 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: $[A] = [A]_{0}*e^{-kt}$? 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.
P: 26
 Quote by mycotheology 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: $[A] = [A]_{0}*e^{-kt}$? 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)
u got your exponentialtion wrong.
P: 392
 Quote by mycotheology 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: $[A] = [A]_{0}*e^{-kt}$? 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*}
##

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