To determine how long it will take for the tranquilizer to decay to 89% of its original dosage, we can use the exponential decay model:
[tex]\[ A = A_0 e^{kt} \][/tex]
Here:
- [tex]\( A_0 \)[/tex] is the original amount of the tranquilizer.
- [tex]\( A \)[/tex] is the remaining amount after time [tex]\( t \)[/tex].
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time.
Given:
- The half-life of the tranquilizer is 34 hours.
- The remaining percentage of the tranquilizer is 89% of the original dosage, which translates to [tex]\( A = 0.89 A_0 \)[/tex].
### Step 1: Determine the decay constant [tex]\( k \)[/tex]
At the half-life, the remaining amount is half of the original amount ([tex]\( A = \frac{A_0}{2} \)[/tex]) after 34 hours. Thus:
[tex]\[ \frac{A_0}{2} = A_0 e^{k \times 34} \][/tex]
Divide both sides by [tex]\( A_0 \)[/tex]:
[tex]\[ \frac{1}{2} = e^{k \times 34} \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln\left(\frac{1}{2}\right) = k \times 34 \][/tex]
Since [tex]\( \ln\left(\frac{1}{2}\right) = \ln(1) - \ln(2) = -\ln(2) \)[/tex], we have:
[tex]\[ k \times 34 = -\ln(2) \][/tex]
Thus,
[tex]\[ k = \frac{-\ln(2)}{34} \][/tex]
### Step 2: Determine the time [tex]\( t \)[/tex] for the drug to decay to 89% of the original dosage
We now need to find [tex]\( t \)[/tex] for [tex]\( A = 0.89 A_0 \)[/tex]:
[tex]\[ 0.89 A_0 = A_0 e^{k \times t} \][/tex]
Divide both sides by [tex]\( A_0 \)[/tex]:
[tex]\[ 0.89 = e^{k \times t} \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.89) = k \times t \][/tex]
Substitute [tex]\( k \)[/tex] from Step 1:
[tex]\[ \ln(0.89) = \left(\frac{-\ln(2)}{34}\right) \times t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.89)}{\frac{-\ln(2)}{34}} \][/tex]
### Step 3: Simplify and find the numerical value
[tex]\[ t = \frac{\ln(0.89) \times 34}{-\ln(2)} \][/tex]
### Step 4: Calculate
Using a calculator for the logarithms:
- [tex]\( \ln(0.89) \approx -0.1165 \)[/tex]
- [tex]\( \ln(2) \approx 0.6931 \)[/tex]
Therefore,
[tex]\[ t = \frac{-0.1165 \times 34}{-0.6931} \][/tex]
[tex]\[ t \approx \frac{-3.961}{-0.6931} \][/tex]
[tex]\[ t \approx 5.716 \][/tex]
### Step 5: Round to the nearest tenth
Rounding to one decimal place:
[tex]\[ t \approx 5.7 \][/tex]
So, it will take approximately [tex]\( \boxed{5.7} \)[/tex] hours for the drug to decay to 89% of its original dosage.
