Radioactivity formula using differential equations? After $12$ days, a $40$ gram substance decays to $9.3$ grams. If the decay rate is proportional to the amount of substance left, how much of the substance will be left after $37$ days?
I solved it using the formula $y=ce^{-kt}$. I used $-k$ instead of $k$ because it is a decay function. I found $k$ by plugging in $(12, 9.3)$, then used that $k$ to find $y$ when $t = 37$. For -k, I got -ln(9.3/40), and so the equation would be y=40e^(-37(ln(9.3/40)/12)). However I got $89$ for y, which was incorrect, so can someone please point out where I made a mistake? Thank you in advance!
 A: There is some confusion with that negative sign in the exponent.
If you are using $-k$ as the coefficient in the exponent, then you are expecting $k$ itself to be positive. You say you found $-k$ to be $-\ln(9.3/40)$, which would mean $k$ was $\ln(9.3/40)$. But $\ln(9.3/40)$ is negative, despite the lack of a negative sign in that expression. Really, $-k=\ln(9.3/40)$, so where you wrote
$$y=40e^{-37(\ln(9.3/40)/12)}$$
it should be
$$y=40e^{37(\ln(9.3/40)/12)}\approx0.445$$

Also you say you got $89$ for $y$. This is (very close to) what I get if I put that negative sign back in and omit the $40$ at the front, so two issues:
$$e^{-37(\ln(9.3/40)/12)}\approx89.85$$
Maybe this is what happened. Maybe with a little rounding error too.
A: $$y=ce^{-kt}$$
The constant $k$ is not correct:
$$ 9.3=40e^{-12k}$$
$$k=-\dfrac 1 {12}\ln \left( \dfrac  {9.3}{40} \right)$$
So that you have:
$$y=40e^{-37k}$$
$$y=40e^{\dfrac {37} {12}\ln \left( \dfrac  {9.3}{40} \right)}$$
$$y=40\left( \dfrac  {9.3}{40} \right)^{\dfrac {37} {12}}$$
A: $$y=40(e^{-kt})\\ 
k=\frac{1}{12}\ln\left(\frac{40}{9.3}\right)\\ 
(y) _{t=37}=40(e^{-\frac{1}{12}\ln\left(\frac{40}{9.3}\right) 37})\\ 
y=40\left(\frac{9.3}{40}\right) ^{\frac{37}{12}}$$
My calculations give $y=0.445$.
