University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 1 - Section 1.5 - Exponential Functions - Exercises - Page 38: 36

Answer

(a) It takes 10 years to reduce the number of cases to 1000. (b) It takes 42 years to eliminate the disease.

Work Step by Step

In any given year, the number of cases of a disease is reduced by $20\%$. So, in any given year, if we take the number of cases of that disease to be $a$, then in the next year, the number of cases would be $80\%$ of $a$, or $\frac{4}{5}a$. Therefore, we can come up with a model to calculate the number of cases of that disease in any year after the first year mentioned: $$a=a_i\Big(\frac{4}{5}\Big)^t$$ $a_i$: the number of cases in the first year mentioned $t$: the number of years that has passed $a$: the number of cases after $t$ years (a) $a_i=10000$ and $a=1000$ Therefore: $$10000\Big(\frac{4}{5}\Big)^t=1000$$ $$\Big(\frac{4}{5}\Big)^t=0.1$$ Using graphing calculator, $$t\approx10.319\approx10 (years)$$ (b) $a_i=10000$ and $a=1$ To have the number of cases less than $1$ would mean: $$10000\Big(\frac{4}{5}\Big)^t\lt1$$ $$\Big(\frac{4}{5}\Big)^t\lt10^{-4}$$ Using graphing calculator, $$t\gt41.275\gt42(years)$$ (Because $\frac{4}{5}\lt1$, we need to change $\lt$ into $\gt$) So it takes 42 years to eliminate the disease.
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