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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 7, Problem 7.RE.21

Caffeine An adult consumes an espresso containing 75 mg of caffeine. If the caffeine has a half-life of 5.5 hours, when will the amount of caffeine in her bloodstream equal 30 mg?

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Identify the exponential decay model for the caffeine amount: \(A(t) = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{h}}\), where \(A(t)\) is the amount of caffeine at time \(t\), \(A_0\) is the initial amount, and \(h\) is the half-life.
Substitute the known values into the model: initial amount \(A_0 = 75\) mg, half-life \(h = 5.5\) hours, and the amount at time \(t\) is \(A(t) = 30\) mg.
Set up the equation: \(30 = 75 \times \left(\frac{1}{2}\right)^{\frac{t}{5.5}}\) to find the time \(t\) when the caffeine amount reaches 30 mg.
Divide both sides by 75 to isolate the exponential term: \(\frac{30}{75} = \left(\frac{1}{2}\right)^{\frac{t}{5.5}}\).
Take the natural logarithm of both sides to solve for \(t\): \(\ln\left(\frac{30}{75}\right) = \frac{t}{5.5} \ln\left(\frac{1}{2}\right)\), then solve for \(t\) by multiplying both sides by 5.5.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential Decay

Exponential decay describes processes where quantities decrease at rates proportional to their current value. In this context, caffeine concentration decreases over time following an exponential decay model, which can be expressed as A(t) = A_0 * (1/2)^(t / half-life).
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Half-Life

Half-life is the time required for a substance to reduce to half its initial amount. For caffeine, a half-life of 5.5 hours means every 5.5 hours, the caffeine amount halves, which is crucial for modeling how long it takes to reach a specific concentration.
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Solving Exponential Equations

To find the time when caffeine reaches 30 mg, one must solve an exponential equation involving the half-life formula. This typically involves isolating the variable in the exponent and using logarithms to solve for time.
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