(a) Find the mass that remains after t years?
t = 30
$P_o=100$
The equation for decay is $P_oe^{kt}$
How do I find the relative growth rate?
So after realizing that the half the amount of 100mg would be 50mg I placed the 50 mg in my formula as the result. Thus,
$50 mg=100e^{k30}$
I divided both sides by 100:
$\frac{50}{100}=\frac{100e^{k30}}{100}$
$ln\ 0.5=ln\ e^{k30}$
$\frac{ln\ 0.5}{30}=k$
k= -0.023
Answering part (b) should be easy after I find part (a)
(b) How much of the sample remains after 100 years?
$P(100)=100e^{-0.023\cdot 100}$
$10.025\ mg=100e^{-0.023\cdot 100}$
(c) After how long will only 1 mg remain?
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$\begingroup$Your answer to a) is correct. Now it is really easy to solve b) and c) becuase you know the decay equation: $$ P(t)=P_0 e^{-0.023 t} $$
so, for b) you have: $$ P(100)=100 e^{-2.3} \approx 10 \mbox{ mg} $$ and for c) you have to solve: $$ 1=100 e^{-0.023 t} $$ and you find: $$ t=-\dfrac{\ln{0.01}}{0.023} \approx 199.3 \mbox{ years} $$
$\endgroup$ $\begingroup$In a, you need to give the units of $k$, but are otherwise correct. For b, it is not rounded correctly and not given as an answer, but OK. For c, use the same equation as you have, $1 mg=100 mg e^{-0.23t}$ and solve for $t$
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