Logarithmic decay

**mathshelpneeded** · January 11th 2007, 10:42 AM

Hello please help!

The rate at which material decays is given in the differential equation where

x is the grams present after t days.

dx/dt = -kx

The half life of a radioactive material is roughly 25 days.

find the time taken for 100g of the material to decay to 20g.

I know you intergrate the equation by swapping x over to the left to get 1/x and dt over to the right to get -kdt. and i integrated the two sides, but when i "plugged" all the numbers in, it never worked out. the answer is...58 days, please help!

**topsquark** · January 11th 2007, 12:05 PM

Originally Posted by mathshelpneeded

Hello please help!

The rate at which material decays is given in the differential equation where

x is the grams present after t days.

dx/dt = -kx

The half life of a radioactive material is roughly 25 days.

find the time taken for 100g of the material to decay.

I know you intergrate the equation by swapping x over to the left to get 1/x and dt over to the right to get -kdt. and i integrated the two sides, but when i "plugged" all the numbers in, it never worked out. the answer is...58 days, please help!

The solution to the differential equation is
$x = x_0e^{-kt}$
where $x_0$ is the initial amount of substance and x is the amount of substance left at time t.

What troubles me about this question is that the answer depends on how much of the material was present initially. Let me explain. We wish to find T such that after T days the amount of material is 100 g less than it was, so $x(T) = x_0 -100$ . Thus:
$x_0 - 100 = x_0e^{-kT}$

Which eventually has the solution:
$T = -\frac{1}{k} \cdot ln \left ( \frac{x_0 - 100}{x_0} \right )$

So the time for 100 g to decay depends on $x_0$ . (In fact, if we only have 100 g to start with, then the amount of time is infinite, as you can verify.)

-Dan

**mathshelpneeded** · January 11th 2007, 12:16 PM

sorry i miswrote the question. the initial mass is 100g, and it is decaying down to 20g, how long will it take?

thanks a lot for helping so speedily!!

**galactus** · January 11th 2007, 12:34 PM

Since you know the 1/2 half, you can use $T=\frac{-1}{k}ln(2)$ to find k.

$25=\frac{-1}{k}ln(2)\Rightarrow{k\approx{-0.027726}}$

$20=100e^{-0.027726t}$

$\frac{1}{5}=e^{-0.027726t}$

$\frac{ln(\frac{1}{5})}{-0.027726}=t=58.05 \;\ days$

**mathshelpneeded** · January 11th 2007, 01:48 PM

Thanks a lot for your help.

just wondering, where did you get the ln(2) from?

**galactus** · January 11th 2007, 02:10 PM

That's the formula for halving time. If you want to see it derived, we may be

able to manage that.

Let $y=y_{0}e^{kt}$ with $y=y_{1}$ when $<br /> <br /> t=t_{1}$ and $y=\frac{y_{1}}{2}$ when

$t=t_{1}+T$ , then $\underbrace{y_{0}e^{kt_{1}}=y_{1}}_{\text{[1]}}$ and $\underbrace{y_{0}e^{k(t_{1}+T)}=\frac{y_{1}}{2}}_{ \text{[2]}}$ . Divide [2] by [1] and get

$e^{-kT}=2, \;\ T=\frac{-1}{k}ln(2)$

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