Question

The​ half-life of a certain tranquilizer in the bloodstream is 47 hours. How long will it take for the drug to decay to 93​% of the original​ dosage? Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve.

Answer 1

Answer:

It will take 4.84 hours for the drug to decay to 93​% of the original​ dosage.

Step-by-step explanation:

We are given that the half-life of a certain tranquilizer in the bloodstream is 47 hours.

The given exponential model is: $A = A_0 e^{kt}$

Now, we know that A becomes half after 47 hours which means that;

$A = 0.5 A_0$

Using this in the above equation we get;

$A = A_0 e^{kt}$

$0.5 A_0 = A_0 e^{(k\times 47)}$  where t = 47 hours

$\frac{0.5 A_0}{A_0} = e^{(47k)}$

$0.5 = e^{47k}$

Taking log on both sides we get;

$ln(0.5) = ln(e^{47k})$

$ln(0.5) =47k$

$k = \frac{ln(0.5)}{47}$

k = -0.015

Now, the time it will take for the drug to decay to 93​% of the original​ dosage is given by;

$0.93 = e^{kt}$  where t is the required time

$0.93 = e^{(-0.015 \times t)}$

Taking log on both sides we get;

$ln(0.93) = ln(e^{-0.015t})$

$ln(0.93) =-0.015t$

$t = \frac{ln(0.93)}{-0.015}$

t = 4.84 hours

Hence, it will take 4.84 hours for the drug to decay to 93​% of the original​ dosage.