A drug is eliminated from the body through urine. Suppose that for a dose of 10 milligrams, the amount A(t) remaining in the body t hours later is given by A(t)=10(0.8)^t and that in order for the drug to be effective, at least 2 milligrams must be in the body.

(a) Determine when 2 milligrams is left in the body.

(b) What is the half-life of the drug?

Thank you!

a)

A(t)=A0(0.8)^t
For 2 mg left, we have A(t)=2, A0=10, so
2=10(0.8)^t
0.8^t = 1/5
take ln on both sides,
t*ln(0.8)=ln(1/5)
t=ln(0.2)/ln(0.8)
=7.2 hours

b)
Half-life is the time it takes to reduce the quantity to 0.5A0, i.e.
5=10(0.8)^t
solve for t similar to a) above.

You're welcome!

To determine when 2 milligrams is left in the body, we need to find the value of t that satisfies the equation A(t) = 2.

(a) Substituting A(t) = 2 into the given equation A(t) = 10(0.8)^t, we get:

2 = 10(0.8)^t

Dividing both sides by 10, we have:

0.2 = (0.8)^t

To solve for t, we can take the logarithm of both sides using the base 0.8:

log(0.2) = log[(0.8)^t]

Using the logarithmic property log(a^b) = b * log(a), we can rewrite the equation as:

log(0.2) = t * log(0.8)

Now, we can divide both sides by log(0.8) to isolate t:

t = log(0.2) / log(0.8)

Using a calculator, we find:

t ≈ 8.66

So, approximately 8.66 hours after a 10 milligram dose, only 2 milligrams of the drug will be left in the body.

(b) The half-life of a substance is the time it takes for half of the initial amount to decay or be eliminated. In this case, the initial amount is 10 milligrams, and we need to find the time it takes for the amount A(t) to reduce to 5 milligrams.

Setting A(t) = 5 in the given equation A(t) = 10(0.8)^t:

5 = 10(0.8)^t

Dividing both sides by 10:

0.5 = (0.8)^t

Again, taking the logarithm of both sides using the base 0.8:

log(0.5) = t * log(0.8)

Dividing both sides by log(0.8):

t = log(0.5) / log(0.8)

Using a calculator:

t ≈ 2.71

So, the half-life of the drug is approximately 2.71 hours.