# Math Help - Need help logarithmic exc.

1. ## Need help logarithmic exc.

Im stuck on this study question.

It says.

Iodine-135 has a half life of 8 days. How long would it take for 28kg of iodine-135 to decay to 200g?

Could someone explain how this answer was gotten? The unit is logarithmic funtions so I assume that they are involved, but how?

2. Originally Posted by Skor
Im stuck on this study question.

It says.

Iodine-135 has a half life of 8 days. How long would it take for 28kg of iodine-135 to decay to 200g?

Could someone explain how this answer was gotten? The unit is logarithmic funtions so I assume that they are involved, but how?
This an exponential decay problem, so the formula is:
$m(t) = m_0e^{-kt}$

where $m_0$ is the original amount of material and k is a constant equal to $\frac{ln(2)}{t_{1/2}}$ where $t_{1/2}$ is the half-life.

So:
$0.200 = 28 e^{-ln(2)t/(8)}$ where t is in days.

$\frac{0.200}{28} = e^{-0.086643t}$

$0.007143 = e^{-0.086643t}$

$ln(0.007143) = -0.086643t$

$-4.94164 = -0.086643t$

$t = \frac{-4.94164}{-0.086643} = 57.0343\, days$

-Dan

3. Hello, Skor!

Iodine-135 has a half-life of 8 days.
How long would it take for 28 kg of Iodine-135 to decay to 200 g?

There is a half-life formula, but we can derive it ourselves.

Assume an exponential function: $I\:=\:c\,e^{-kt}$
. . where $I$ is the amount of I-135 at time $t$ (days).
Let $I_o$ = original amount of I-135.

At the very beginning $(t = 0)$, we have: . $I_o \:=\:c\,e^0\quad\Rightarrow\quad c\:=\:I_o$

In 8 days, only half of the I-135 remains. .That is, when $t = 8,\;I = \frac{1}{2}I_o$
So we have: . $\frac{1}{2}I_o\:=\:I_o\,e^{-8k}\quad\Rightarrow\quad e^{-8k} \:=\:\frac{1}{2}$

Rewrite in logarithmic form: . $-8k \:=\:\ln\left(\frac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \:=\:-\ln 2$
. . Hence:. . $k \:=\:\frac{\ln 2}{8}$

Therefore, the function is: . $I \:=\:I_o\,e^{-\frac{\ln2}{8}t}$

We are told that: . $I_o \:=\:28\text{ kg}\:=\:28,000\text{ g}$ and $I \:=\:200\text{ g}$

So we have: . $200 \:=\:28,000e^{-\frac{\ln 2}{8}t}$

. . $e^{-\frac{\ln2}{8}t}\:=\:\frac{1}{140}\quad\Rightarrow \quad -\frac{\ln 2}{8}t \:=\:\ln\left(\frac{1}{140}\right) \:=\:-\ln140$

Therefore: . $t\:=\:\frac{8\ln140}{\ln2} \:=\:57.03426414\:\approx\: 57.03\text{ days}$