University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 81


(a) If we call the remaining amount of substance after $t$ hours $y_0$: $$y_0=8\times\Big(\frac{1}{2}\Big)^{t/12}$$ (b) There will be 1 gram remaining after $36$ hours.

Work Step by Step

(a) The half-life of the radioactive substance is $12$ hours means that after $12$ hours, there remains half the original amount of that substance. For example, here we have $8$ grams initially. After $12$ hours, what we have left is $8\times(\frac{1}{2})=8\times(\frac{1}{2})^{12/12}=4$ grams. After $16$ hours, the remaining amount is $8\times(\frac{1}{2})^{16/12}$ grams and so on. Therefore, if we call the remaining amount of substance after $t$ hours $y_0$, we can come up with the model to calculate that amount: $$y_0=8\times\Big(\frac{1}{2}\Big)^{t/12}$$ (b) We know here the remaining amount is $1$ gram, meaning $y_0=1$. Therefore, $$8\times\Big(\frac{1}{2}\Big)^{t/12}=1$$ $$\Big(\frac{1}{2}\Big)^{t/12}=\frac{1}{8}=\frac{1}{2^3}=\Big(\frac{1}{2}\Big)^3$$ $$\frac{t}{12}=3$$ $$t=36(hours)$$
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