# decay problem

• Mar 3rd 2010, 09:27 PM
break
decay problem
A substance decays at a rate proportional to the amount present, and half of the original quantity Yo is left after 1000 years. What is the percent of the original quantity Yo left after 2000 years?

basically i use the formula
Y = Yo*e^(-kt)

then i solve for k

$\displaystyle \dfrac{1}{2} = 1 * e^{-1000k}$

$\displaystyle \dfrac{\ln 1}{\ln 2} = \ln e^{-1000k}$

$\displaystyle \dfrac{\ln 1}{\ln 2} = -1000k$

$\displaystyle k = \dfrac{\ln \dfrac{1}{2}}{-1000}$

After i get k, then i solve for Y.

$\displaystyle Y = 1e^{(2000)* -(\dfrac{\ln \dfrac{1}{2}}{-1000})}$

is this correct?
• Mar 3rd 2010, 09:45 PM
Gojinn
You can simplify this down further.

$\displaystyle \ln\dfrac{1}{2} \equiv \ln 1 - \ln 2$

$\displaystyle \ln1 = 0$

So now you have:

$\displaystyle -\ln2 = -1000k$

$\displaystyle \ln2 = 1000k$

$\displaystyle k = \dfrac{\ln2}{1000}$

This makes working out the second part of the question easier. As you've stated:

$\displaystyle Y = e^{(2000)* -\dfrac{\ln2}{1000}}$

$\displaystyle Y = e^{-2\ln2}$

$\displaystyle Y = (e^{\ln2})^{-2}$

$\displaystyle Y = (2)^{-2}$

$\displaystyle Y = \dfrac{1}{4}$
• Mar 3rd 2010, 09:53 PM
break
is that all thats different?

so

$\displaystyle Y = 1e^{(2000)* (\dfrac{-\ln 2}{1000})}$
$\displaystyle Y = .25$

After 2000 years, quantity left is 25%.

correct?
• Mar 3rd 2010, 09:58 PM
Gojinn
Yeah, I forgot to follow up in my post. lol

Yeah, .25 is what I got as well.
• Mar 3rd 2010, 10:05 PM
Soroban
Hello, break!

Quote:

A substance decays at a rate proportional to the amount present,
and half of the original quantity Yo is left after 1000 years.
What is the percent of the original quantity Yo left after 2000 years?

Basically i use the formula: .$\displaystyle Y \:=\: Y_o\,e^{-kt}$

then i solve for $\displaystyle k\!: \;\;\frac{1}{2} \:=\:e^{-1000k}$

. . $\displaystyle \frac{\ln 1}{\ln 2} = \ln e^{-1000k}$

. . .$\displaystyle {\color{blue}\uparrow}$
. wrong!

We have: .$\displaystyle e^{-1000k} \:=\:\frac{1}{2}$

Then: .$\displaystyle \ln\left(e^{-1000k}\right) \:=\:\ln\left(\frac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \;=\;-\ln(2)$

$\displaystyle \text{Hence: }\;-1000k\underbrace{\ln(e)}_{\text{This is 1}} \:=\:-\ln(2) \quad\Rightarrow\quad -1000k \:=\:\-\ln(2) \quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{1000}$

Therefore, the function is: .$\displaystyle Y \;=\;Y_o\,e^{-\frac{\ln(2)}{1000}t}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The answer can be further simplified . . .

. . $\displaystyle Y \;=\;Y_o\,e^{-\frac{\ln(2)}{1000}t} \;=\;Y_o\left(e^{\ln(2)}\right)^{-\frac{t}{1000}} \quad\Rightarrow\quad Y \;=\;Y_o\cdot2^{-\frac{t}{1000}}$

• Mar 3rd 2010, 10:16 PM
Gojinn
Quote:

$\displaystyle \;-1000k\underbrace{\ln(e)}_{\text{This is 1}} \:=\:-\ln(2) \quad\Rightarrow\quad -1000k \:=\:\-\ln(2) \quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{1000}$
I think you made a mistake. Shouldn't it state:

$\displaystyle \;-1000k\underbrace{\ln(e)}_{\text{This is 1}} \:=\:-\ln(2) \quad\Rightarrow\quad -1000k \:=\:-\ln(2) \quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{1000}$

Only a small thing, but just makes it clearer. lol.