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The half-life of a certain tranquilizer in the bloodstream is 34 hours. How long will it take for the drug to decay to [tex]89 \%[/tex] of the original dosage?

Use the exponential decay model, [tex]A = A_0 e^{kt}[/tex], to solve.

[tex]\(\square\)[/tex] hours

(Round to one decimal place as needed.)

Sagot :

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.