Question

the half - life of a certain tranquilizer in the bloodstream is 22 hours. how long will it take for the drug to decay to 91% 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: Find the decay constant $k$

The half - life formula is $A = A_0e^{kt}$. When $t = t_{1/2}$ (half - life), $A=\frac{A_0}{2}$. So, $\frac{A_0}{2}=A_0e^{kt_{1/2}}$. Divide both sides by $A_0$ to get $\frac{1}{2}=e^{kt_{1/2}}$. Given $t_{1/2}=22$ hours, then $\ln(\frac{1}{2})=\ln(e^{22k})$. Since $\ln(e^{x}) = x$, we have $- \ln(2)=22k$. Solving for $k$, we get $k=\frac{-\ln(2)}{22}\approx - 0.0315$.

Step2: Find the time $t$ when $A = 0.91A_0$

Substitute $A = 0.91A_0$ and $k\approx - 0.0315$ into the decay formula $A = A_0e^{kt}$. Divide both sides by $A_0$: $0.91=e^{- 0.0315t}$. Take the natural logarithm of both sides: $\ln(0.91)=\ln(e^{-0.0315t})$. Since $\ln(e^{x}) = x$, we have $\ln(0.91)=-0.0315t$. Then $t=\frac{\ln(0.91)}{- 0.0315}$.

Step3: Calculate the value of $t$

We know that $\ln(0.91)\approx - 0.0943$. So, $t=\frac{-0.0943}{-0.0315}\approx3.0$ hours.

Answer:

$3.0$ hours