Question

the half - life of a certain tranquilizer in the bloodstream is 49 hours. how long will it take for the drug to decay to 92% of the original dosage? use the exponential decay model, $a = a_0e^{kt}$, to solve. (round the final answer to one decimal place as needed. round all intermediate values to four decimal places as needed.)

Explanation:

Step1: Recall half - life formula

For exponential decay model \(A = A_0e^{kt}\), when \(t = t_{1/2}\) (half - life), \(A=\frac{A_0}{2}\). So \(\frac{A_0}{2}=A_0e^{kt_{1/2}}\). We can cancel \(A_0\) (since \(A_0
eq0\)) and get \(\frac{1}{2}=e^{kt_{1/2}}\). Given \(t_{1/2} = 49\) hours, we solve for \(k\):
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{49k})\). Since \(\ln(e^{x})=x\), we have \(\ln(\frac{1}{2}) = 49k\). Then \(k=\frac{\ln(0.5)}{49}\approx\frac{- 0.693147}{49}\approx - 0.014146\).

Step2: Set up the equation for 92% decay

We want the drug to decay to 92% of the original dosage, so \(A=(1 - 0.08)A_0=0.92A_0\). Using the exponential decay model \(A = A_0e^{kt}\), we substitute \(A = 0.92A_0\) and \(k=-0.014146\):
\(0.92A_0=A_0e^{-0.014146t}\)
Cancel \(A_0\) (since \(A_0
eq0\)): \(0.92 = e^{-0.014146t}\)

Step3: Solve for \(t\)

Take the natural logarithm of both sides: \(\ln(0.92)=\ln(e^{-0.014146t})\)
Since \(\ln(e^{x}) = x\), we have \(\ln(0.92)=- 0.014146t\)
Then \(t=\frac{\ln(0.92)}{- 0.014146}\). Calculate \(\ln(0.92)\approx - 0.083291\)
So \(t=\frac{-0.083291}{-0.014146}\approx5.8877\)

Answer:

\(5.9\) (rounded to one decimal place)