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

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2 years ago