Chapter 0.3, Problem 32E

a.

To determine

To Reduce: The number of cases to $1000$ .

Answer to Problem 32E

About $10.3$ years will take to reduce the number of cases to $1000$ .

Explanation of Solution

Given:

Number of cases of a disease is reduced by $=20\%$ .

If there are $10,000$ cases today, find number of years will it take to the number of cases to $1000$ .

Calculation:

As a result of a $20\%$ reduction in cases, the number is multiplied by $0.8$ . Thus, after $t$ years

there will be:

  $y=10,000\cdot {0.8}^t$

Substitute $y=1000$ :

  $\begin{array}{c}y=10,000\cdot {0.8}^t\\ 1000=10,000\cdot {0.8}^t\\ \frac{1000}{10,000}={0.8}^t\\ 0.1={0.8}^t\end{array}$

Take natural logarithm in both sides,

  $\begin{array}{c}\ln 0.1=t\ln 0.8\\ \frac{\ln 0.1}{\ln 0.8}=t\\ 10.3\text{ years}=t\end{array}$

It will take $10.3$ years to reduce the number of cases to $1000$ .

b.

To determine

To Reduce: The number of cases to less than $1$ . And find the year the disease will completely eradicate.

Answer to Problem 32E

Tt will take $41.3$ years to reduce the number of cases to $1$ . And, it will take $42$ years to completely eradicate the disease.

Explanation of Solution

Given:

Number of cases of a disease is reduced by $=20\%$ .

If there are $10,000$ cases today, find number of years will it take to the number of cases to $1000$ .

Calculation:

From part $\left(a\right)$ , we have

  $y=10,000\cdot {0.8}^t$

Substitute $y=1$ :

  $1=10,000\cdot {0.8}^t$

Rewrite it as:

  $\begin{array}{c}10,000\cdot {0.8}^t=1\\ {0.8}^t=0.0001\end{array}$

Take logarithm on both sides,

  $\begin{array}{c}t\log 0.8=\log 0.0001\\ t\log \left(8\cdot {10}^{-1}\right)=\log {10}^{-4}\\ t(\log 8-1)=-4\\ t=\frac{-4}{\log 8-1}\\ t\approx 41.3\end{array}$

So, it will take $41.3$ years to reduce the number of cases to $1$ . Thus, it will take $42$ years to completely eradicate the disease.