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The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.The half-life of the plutonium isotope is 24,360 years. If $10 \mathrm{g}$ of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for $80 \%$ of the isotope to decay?

56,562 years

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

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we're told that the plutonium isotope has 1/2 life of 24,360 years. Yeah, 10 grams of plutonium released into the atmosphere will determine how long will it take before 80% of the isotope has decayed? No. They tell us in this chapter that whenever we're talking about radioactive decay that Paul's an exponential growth model. So are amount will be something based off. Why? T is eager to. Why not eat the negative? Katie, Like this here and we can go and solve for Kay, actually, pretty quickly because they tells that half life which I write is t sub 1/2. Does he go to the natural log too? Over K. So if we want to solve Okay, we just need to do K is equal to the natural log to over or half life. Sorry. So we would just plug in at 24,360 jokes natural August 2, Boger 24,360. And the units for this would be per year. Now we can plug all that in to help us soul. For this now, 80% decay means 20% left and We don't even need to know how much we have initially to start. Because if we write a 0.2, why not? For how much we have left is equal to Why not our initial and then e to the I'm just gonna leave its cave right now. Negative, K. T. Um, when we divide by, Why not? Those just counts out each other. So it doesn't even matter all that much right now. Let's solve for tea. And then after that, we can plug in for Kay. So we would take the natural log on each side and we get natural log 0.2 zero to negative, Katie. Now let's divide by negative K on each side so we'll get T is equal to the natural log of 0.2 over negative K. All right, now, let's plug in what k is. So this is K. So since it's one over Kay, we would just reciprocate it. So it should be negative. Natural log of 0.2 times 24,360 over the natural log of two. So this would be in years since Kay was reciprocal years. And let's see what this is going to be approximately equal to negative. Natural log of 0.2 is about 1.609 times 24,360 is about 39,000. And then divide that by natural log two and this is going to give somewhere around 56,000 562 years. So this is how long it will take or 80% to decay.

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