Hint: If you start with 1 unit of the isotope at time $0$, you have $\frac 12$ after $150$ years. Do you know how to solve $m'=km$? That should give you an equation that (with this data) lets you determine $k$. Now put that $k$ into your solution and find the time when the amount is $0.15$.
Added: your equation is the correct differential equation. The solution is $m(t)=m(0)\exp(-kt)$ as the derivative of $\exp(-kt)$ is $-k \exp(-kt)$
I will get you kicked off. As mentioned in the comments, we use an exponential for these types of problems, so we have:
$$A(t) = ce^{kt}$$
Now all you need to do is use the conditions specified in the problem to find the particular solution.
- Using $t = 30$ and $A(30) = 100$, we get: $\displaystyle 100 = ce^{30k}$.
- Using $t = 120$ and $A(120) = 30$, we get: $\displaystyle 30 = ce^{120k}$.
The resulting system of equations is:
$$100 = ce^{30k}$$
$$30 = ce^{120k}$$
Using the first equation, we get:
$$\displaystyle c = \frac{100}{e^{30k}} = 100e^{-30k}$$
Substituting into the second equation, you have: $30 = 100e^{-30k} e^{120k} = 100e^{90k}$, or, $\frac{30}{100} = e^{90k}$, so
$$\displaystyle \ln\left(\frac{30}{100}\right) = 90k \rightarrow k = -0.013377$$
Notice here that $k$ is in fact negative. This is what you would expect given that we are talking about decay rather than growth.
Using the equation for $c$, we have:
$$\displaystyle c = 100e^{-30 \times -0.013377} = 149.38$$
The amount $A$ as a function of time is given by:
$$\displaystyle A(t) = 149.38e^{-0.013377t}$$
To find the original amount of substance, use $t=0$, and we get:
$$A(0) = 149.38e^0 = 149.38~~ \text{mg}$$
How long will it take before only 1% of the "original amount remains"?
$$149.38 e^{-0.013377t} = .01 \times 149.38 \rightarrow t = 344.26 ~~ \text{days}.$$
Best Answer
C-14 is just the name of the isotope it is not an amount of the substance.
The amount at any time is given by,
$$r(t) = Ce^{\lambda t}$$
The amount at a later time $t+\tau$ should be half as much (\tau is the half life).
$$\frac{1}{2} r(t)= r(t + \tau) = C e^{\lambda (t+\tau)} = C e^{\lambda t} e^{\lambda \tau} = r(t) e^{\lambda \tau} $$
From this we conclude that,
$$ \frac{1}{2} = e^{\lambda \tau}$$
$$ \frac{\ln(1/2)}{ \tau} = \lambda $$