Dayaan M. answered • 08/26/26
Scored 5/5 on Algebra 2 EOC | 5 Years of Tutoring Experience
In order to work with half-life, the most useful form of the exponential decay equation is the one that has the half-life built right into it:
A(t) = A0 (1/2)^(t/h)
Here A0 is how much you start with, h is the half-life, t is time, and A(t) is how much is left. The reason this form is so convenient is that you can read it in plain English. Every time t reaches another full h, the exponent goes up by 1 and you cut what is left in half one more time.
Filling in what we know, A0 is 30 mg, h is 5.7 hours, and we want to know when A(t) reaches 5 mg:
5 = 30(1/2)^(t/5.7)
First we get the exponential piece by itself by dividing both sides by 30:
1/6 = (1/2)^(t/5.7)
Now t is stuck in the exponent, so we take the natural log of both sides. The power rule for logs lets the exponent come down in front, which is exactly what we need:
ln(1/6) = (t/5.7) ln(1/2)
Dividing both sides by ln(1/2):
t/5.7 = ln(1/6)/ln(1/2)
Both of those logs are negative, so the negatives cancel and we get a positive number, about 2.585. Then we multiply both sides by 5.7:
t = 5.7 x 2.585 = 14.7
Remember, it is always worth sanity checking a decay answer. After one half-life at 5.7 hours we would be at 15 mg, after two at 11.4 hours we would be at 7.5 mg, and after three at 17.1 hours we would be down to 3.75 mg. Our answer of 14.7 hours falls between the second and third of those, and 5 mg falls between 7.5 and 3.75, so it lines up.
So, our final answer is that it takes about 14.7 hours.
