Amount left = initial amount * e^(rate * time)

1 = 3 * e^ (rate * 4)

Let's solve for the rate:

1/3 = e^(4r)

ln(1/3) = 4r

(1/4)ln(1/3) = r

Solve for time in this equation:a) How long will it take to have 2 grams left?

2 = 3 e^(rt) // we solved for r above, plug it in and isolate t

Solve for y:b) How much will you have after 6 seconds?

y = 3 e^(6r) // again we found r above

Solve for t:c) What is the half-life of the substance?

(3/2) = 3 e^(rt) // once more, we already know r.

These styles of problem have 4 variables: final amount, initial amount, rate, and time. Arrange a problem that plugs in 3 things, and find the 4th.