We want to see how long will take a healthy adult to reduce the caffeine in his body to a 60%. We will find that the answer is 3.55 hours.
We know that the half-life of caffeine is 4.8 hours, this means that for a given initial quantity of coffee A, after 4.8 hours that quantity reduces to A/2.
So we can define the proportion of coffee that Jeremiah has in his body as:
P(t) = 1*e^{k*t}
Such that:
P(4.8 h) = 0.5 = 1*e^{k*4.8}
Then, if we apply the natural logarithm we get:
Ln(0.5) = Ln(e^{k*4.8})
Ln(0.5) = k*4.8
Ln(0.5)/4.8 = k = -0.144
Then the equation is:
P(t) = 1*e^{-0.144*t}
Now we want to find the time such that the caffeine in his body is the 60% of what he drank that morning, then we must solve:
P(t) = 0.6 = 1*e^{-0.144*t}
Again, we use the natural logarithm:
Ln(0.6) = Ln(e^{-0.144*t})
Ln(0.6) = -0.144*t
Ln(0.6)/-0.144 = t = 3.55
So after 3.55 hours only the 60% of the coffee that he drank that morning will still be in his body.
If you want to learn more, you can read:
brainly.com/question/19599469
