How would I solve the following problem?
A year ago there were 4 gram of a radioactive substance.Now there are 3 grams. How much was there 10 years ago?
I did the following
$4e^{k(1)}=3$ With four grams being my initial grams and this happened 1 year ago.
for my $k$ I got $k=\ln\frac{3}{4}$
Then I set up the equation
$$4e^{\ln\frac{3}{4}(t)}$$
But what should I use for my time or t I tired -9 because 1 year passed and you want to know how many grams you had 10 years ago.
And i go 53.271 grams but that seems wrong.
Best Answer
You are quite close, possibly off a bit only by a copying error or low percision: $$\left(\frac 43 \right)^{10}\cdot 3 = \left(\frac 43 \right)^{9}\cdot 4=\frac{1048576}{19683}\approx 53.27318$$