Question

the half-life of a certain tranquilizer in the bloodstream is 38 hours. how long will it take for the drug to decay to 90% of the original dosage? use the exponential decay model, $a = a_0e^{kt}$, to solve.

\boxed{} hours
(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 \( T_{1/2}=38 \) hours. For exponential decay \( A = A_0e^{kt} \), when \( t = T_{1/2} \), \( A=\frac{A_0}{2} \). So we substitute into the formula:
\(\frac{A_0}{2}=A_0e^{k\times38}\)
Divide both sides by \( A_0 \) (since \( A_0
eq0 \)): \(\frac{1}{2}=e^{38k}\)
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{38k})\)
Using the property \( \ln(e^x)=x \), we get \(\ln(\frac{1}{2}) = 38k\)
Then \( k=\frac{\ln(\frac{1}{2})}{38}=\frac{-\ln(2)}{38}\approx\frac{- 0.6931}{38}\approx - 0.0182\)

Step2: Find the time \( t \) when \( A = 0.9A_0 \)

We use the formula \( A = A_0e^{kt} \), substitute \( A = 0.9A_0 \) and \( k\approx - 0.0182 \):
\(0.9A_0=A_0e^{-0.0182t}\)
Divide both sides by \( A_0 \): \(0.9 = e^{-0.0182t}\)
Take the natural logarithm of both sides: \(\ln(0.9)=\ln(e^{-0.0182t})\)
Using the property \( \ln(e^x)=x \), we get \(\ln(0.9)=- 0.0182t\)
Then \( t=\frac{\ln(0.9)}{- 0.0182}\)
We know that \(\ln(0.9)\approx - 0.1054\), so \( t=\frac{-0.1054}{-0.0182}\approx5.7912\approx5.8\) (rounded to one decimal place)

Answer:

\(5.8\)