- Thread starter
- #1

#### bergausstein

##### Active member

- Jul 30, 2013

- 191

can you me go about solving this. thanks!

- Thread starter bergausstein
- Start date

- Thread starter
- #1

- Jul 30, 2013

- 191

can you me go about solving this. thanks!

- Admin
- #2

- Thread starter
- #3

- Jul 30, 2013

- 191

now the half-life is 1600 so,

$1600=3200e^{kt}$

now how wil I solve for the rate of decline here? please help!

- Admin
- #4

You have the correct equation for the mass of a sample, but the half-life being 1600 years means that at time t=1600 then $R(t)=\dfrac{1}{2}R_0$. That is (I like to always use a positive constant):

now the half-life is 1600 so,

$1600=3200e^{kt}$

now how wil I solve for the rate of decline here? please help!

\(\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0\)

Divide through by \(\displaystyle R_0\) and the convert from exponential to logarithmic form to solve for $k$. Another way to look at it is:

\(\displaystyle R(t)=R_02^{-\frac{t}{1600}}\)

And then to find the percentage remaining after $t$ years, use:

\(\displaystyle \frac{100R(t)}{R_0}=100\cdot2^{-\frac{t}{1600}}\)

- Thread starter
- #5

- Jul 30, 2013

- 191

from here solving for k

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?

Last edited:

- Admin
- #6

- Thread starter
- #7

- Jul 30, 2013

- 191

how can I do that if there's no given initial amount? can you tell how to go about it?You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.

Do you think it's a good idea to give a decimal approximation when the function simplifies greatly if kept exact?from here solving for k

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?

[tex]\displaystyle \begin{align*} k &= \frac{1}{1600} \ln{ (2) } \end{align*}[/tex]

and so

[tex]\displaystyle \begin{align*} R(t) &= R_0 \exp { \left[ \frac{t}{1600} \ln{(2)} \right] } \\ &= R_0 \exp{ \left[ \ln{ \left( 2 ^{ \frac{t}{1600} } \right) } \right] } \\ &= R_0 \cdot 2^{ \frac{t}{1600} } \end{align*}[/tex]

and so after 200 years, what proportion of the initial amount do you have?

- Admin
- #9

This is what I posted before:how can I do that if there's no given initial amount? can you tell how to go about it?

...to find the percentage remaining after $t$ years, use:

\(\displaystyle \frac{100R(t)}{R_0}\)

- Thread starter
- #10

- Jul 30, 2013

- 191

do you mean like this? how can I find the percentage of remaining here?

- Admin
- #11

Reduce the exponent, and then that is the exact percentage remaining, and then you can use a calculator to obtain a decimal approximation if you like.

do you mean like this? how can I find the percentage of remaining here?

You know that your model involves exponential decay, so after each unit of time passes, there is a certain constant amount multiplying through. In other words

[tex]\displaystyle \begin{align*} R_1 &= C\,R_0 \\ R_2 &= C^2\,R_0 \\ R_3 &= C^3\,R_0 \\ \vdots \\ R_t &= C^t\,R_0 \end{align*}[/tex]

Since you know that when [tex]\displaystyle \begin{align*} t = 1600, R_{1600} = \frac{1}{2}R_0 \end{align*}[/tex], that means

[tex]\displaystyle \begin{align*} \frac{1}{2}R_0 &= C^{1600}\,R_0 \\ \frac{1}{2} &= C^{1600} \\ \ln{ \left( \frac{1}{2} \right) } &= \ln{ \left( C^{1600} \right) } \\ \ln{ \left( \frac{1}{2} \right) } &= 1600 \ln{(C)} \\ \frac{1}{1600} \ln{ \left( \frac{1}{2} \right) } &= \ln{(C)} \\ \ln{ \left[ \left( \frac{1}{2} \right) ^{\frac{1}{1600}} \right] } &= \ln{(C)} \\ \left( \frac{1}{2} \right) ^{ \frac{1}{1600} } &= C \\ 2^{ -\frac{1}{1600}} &= C \end{align*}[/tex]

and thus your model is [tex]\displaystyle \begin{align*} R(t) = \left( 2^{-\frac{1}{1600}} \right) ^t \, R_0 = 2^{-\frac{t}{1600}}\,R_0 \end{align*}[/tex].

Now if [tex]\displaystyle \begin{align*} t = 200 \end{align*}[/tex], that gives

[tex]\displaystyle \begin{align*} R(200) &= 2^{-\frac{200}{1600}}\,R_0 \\ &= 2^{-\frac{1}{8}}\,R_0 \\ &\approx 0.917 \, R_0 \end{align*}[/tex]

So what percentage of the original amount do you have?

- Admin
- #13

Yeah, I missed the dropped negative sign in the exponent.It appears there was something wrong with your initial model.