Question

The half-life of a certain radioactive substance is 80 days. There are 4 grams initially. When will there be 1 gram remaining?

Answer 1

Use the standard formula \[A(t)= A_{0}e^{rt}\]

Answer 2

Is this the right set up? \[1= 4e^{80t}\]

Answer 3

the rate of decay is 80?

Answer 4

how did you get that rate ?

Answer 5

We are given by first sentence the following:
\[\frac{1}{2} A_0=A_0e^{r \cdot 80} \\ \text{ solve for } r \\ \text{ where } r \text{ is the rate } \\ t=80 \text{ is the amount of days \it takes for some amount to get half }\]

Answer 6

once you find r you can back to that 1=4*e^(rt)
where you know r now from the previous equation I just asked you to solve

Answer 7

is r= ln(2/4) / 80 ?

Answer 8

2/4 is the same as 1/2

Answer 9

oh ok just reduced

Answer 10

so you can write as that or this last line I have here
\[r=\frac{1}{80} \ln(\frac{1}{2}) \\ r=\frac{1}{80}(\ln(1)-\ln(2)) \\ r=\frac{1}{80}(0-\ln(2)) \\ r=\frac{-\ln(2)}{80}\]

Answer 11

\[1=4 e^{ rt } \\ \text{ where we found } r=\frac{-\ln(2)}{80} \\ 1=4 e^{\frac{-\ln(2)}{80} t} \\ \text{ solve for } t \]

Answer 12

mention me if you have trouble
my notifications aren't workiong

Answer 13

@freckles hey I get t= 160 :}

Answer 14

one sec checking...

Answer 15

sounds great