Question

1. drug concentration the concentration of a drug in a patients bloodstream after t hours is modeled by the formula c(t) = 11(0.72)^t, where c is measured in milligrams per liter. (a) what is the initial concentration of the drug? (b) how long does it take for the concentration to decrease to 50% of its initial level?

Explanation:

Step1: Find initial concentration

Initial concentration is when $t = 0$. Substitute $t=0$ into $C(t)=11(0.72)^t$. Since any non - zero number to the power of 0 is 1, we have $C(0)=11(0.72)^0$.
$C(0)=11\times1 = 11$

Step2: Find time for concentration to decrease to 50%

The initial concentration is 11. 50% of the initial concentration is $0.5\times11 = 5.5$. We set $C(t)=5.5$ in the equation $C(t)=11(0.72)^t$. So, $5.5 = 11(0.72)^t$.
First, divide both sides by 11: $\frac{5.5}{11}=(0.72)^t$, which simplifies to $0.5=(0.72)^t$.
Take the natural logarithm of both sides: $\ln(0.5)=\ln((0.72)^t)$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(0.5)=t\ln(0.72)$.
Then solve for $t$: $t=\frac{\ln(0.5)}{\ln(0.72)}\approx\frac{- 0.6931}{-0.3283}\approx2.11$

Answer:

(a) 11 milligrams per liter
(b) Approximately 2.11 hours