1. exponential growth

Hi, I'm having trouble using the formula for exponential growth when finding the half-life.

Q: A sample of Bismuth-210 decayed to 33 percent of its original mass after eight days.

find the half life of this element and then find the mass remaining after 12 days.

The formula that I am required to use for this question is:
y_t=y_o(e)^(kt)

I cannot simply use the half-life formula aka: y_t=y_o(0.5)^(t/h).

I know for this general exponential growth formula that I must find the value of k, the constant first... and when dealing with the half-life. So here is my work, but it is not right.. as it does not match the answer in the back.

I took half of the initial amount: so.. 100-33=67 therefore 67% left... half of that is 33.5% left so
33.5=100(e)^(8k)
ln0.335=lne^(8k)
k=ln0.335/8

2. Originally Posted by skeske1234
Hi, I'm having trouble using the formula for exponential growth when finding the half-life.

Q: A sample of Bismuth-210 decayed to 33 percent of its original mass after eight days.

find the half life of this element and then find the mass remaining after 12 days.

The formula that I am required to use for this question is:
y_t=y_o(e)^(kt)
$\displaystyle .33 = e^{8k}$

$\displaystyle \ln(.33) = 8k$

$\displaystyle k = \frac{\ln(.33)}{8}$

$\displaystyle .5 = e^{kt}$

$\displaystyle \ln(.5) = kt$

$\displaystyle t = \frac{\ln(.5)}{k} = \frac{8\ln(.5)}{\ln(.33)}$

half life , $\displaystyle t \approx 5$ days

3. how do i find the mass remaining after 12 days? the answer is 19% of original amount......

4. Originally Posted by skeske1234
how do i find the mass remaining after 12 days? the answer is 19% of original amount......
$\displaystyle y = e^{12k}$

with $\displaystyle k = \frac{\ln(.33)}{8}$ ...

$\displaystyle y \approx .19$ ... 19%