In Exercises 19 and 20, find the amount of time required for a $$ \$ 2000 $$ investment to double if the annual interest rate $$ r $$ is compounded (a) annually, (b) monthly, (c) quarterly, and (d) continuously. 19. $$ r=4.75 \% $$ 20. $$ r=8.25 \% $$ 21. Half-Life The radioactive decay of $$ \mathrm{Sm}-151 $$ (an isotope of samarium) can be modeled by the differential equation $$ d y / d t= $$ $$ -0.0077 y $$, where $$ t $$ is measured in years. Find the half-life of Sm-151. 22. Half-Life An isotope of neptunium ( Np -240) has a half-life of 65 minutes. If the decay of $$ \mathrm{Np}-240 $$ is modeled by the differential equation $$ d y / d t=-k y $$, where $$ t $$ is measured in minutes, what is the decay constant $$ k $$ ?

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**19. For $$ r=4.75\% $$ (i.e. $$ r=0.0475 $$)**

We want the amount to double. The general compound‐interest formula is  
$$
A = P\left(1+\frac{r}{n}\right)^{nt}\,.
$$
Setting $$ A=2P $$ (doubling) gives  
$$
2 = \left(1+\frac{r}{n}\right)^{nt}\,.
$$
Taking natural logarithms we have  
$$
\ln 2 = nt \cdot \ln\left(1+\frac{r}{n}\right)
$$
so that  
$$
t=\frac{\ln 2}{n\ln\left(1+\frac{r}{n}\right)}\,.
$$

**(a) Annually ($$ n=1 $$):**  
$$
t=\frac{\ln 2}{\ln(1+0.0475)}\,. 
$$
Numerically, since  
$$
\ln(1.0475)\approx0.04637,
$$
we get  
$$
t\approx\frac{0.69315}{0.04637}\approx14.95\text{ years}\,.
$$

**(b) Monthly ($$ n=12 $$):**  
$$
t=\frac{\ln 2}{12\ln\left(1+\frac{0.0475}{12}\right)}\,.
$$
Here,  
$$
\frac{0.0475}{12}\approx0.003958\,,
$$
and  
$$
\ln(1.003958)\approx0.00395\,.
$$
Thus,  
$$
t\approx\frac{0.69315}{12\times0.00395}\approx\frac{0.69315}{0.0474}\approx14.62\text{ years}\,.
$$

**(c) Quarterly ($$ n=4 $$):**  
$$
t=\frac{\ln 2}{4\ln\left(1+\frac{0.0475}{4}\right)}\,.
$$
We have  
$$
\frac{0.0475}{4}=0.011875 \quad \text{and} \quad \ln(1.011875)\approx0.01181\,.
$$
Then,  
$$
t\approx\frac{0.69315}{4\times0.01181}\approx\frac{0.69315}{0.04724}\approx14.67\text{ years}\,.
$$

**(d) Continuously:**  
For continuous compounding, the formula is  
$$
A=P\, e^{rt}\,.
$$
Setting $$ 2P=P\, e^{rt} $$ gives  
$$
2=e^{rt}\,,
$$
so that  
$$
t=\frac{\ln2}{r}=\frac{0.69315}{0.0475}\approx14.59\text{ years}\,.
$$

---

**20. For $$ r=8.25\% $$ (i.e. $$ r=0.0825 $$)**

Using the same steps with $$ r=0.0825 $$:

**(a) Annually ($$ n=1 $$):**  
$$
t=\frac{\ln2}{\ln(1+0.0825)}\,.
$$
Since  
$$
\ln(1.0825)\approx0.0793,
$$
we obtain  
$$
t\approx\frac{0.69315}{0.0793}\approx8.74\text{ years}\,.
$$

**(b) Monthly ($$ n=12 $$):**  
$$
t=\frac{\ln2}{12\ln\left(1+\frac{0.0825}{12}\right)}\,.
$$
Here,  
$$
\frac{0.0825}{12}\approx0.006875,\quad \ln(1.006875)\approx0.006853.
$$
Thus,  
$$
t\approx\frac{0.69315}{12\times0.006853}\approx\frac{0.69315}{0.082236}\approx8.42\text{ years}\,.
$$

**(c) Quarterly ($$ n=4 $$):**  
$$
t=\frac{\ln2}{4\ln\left(1+\frac{0.0825}{4}\right)}\,.
$$
Now,  
$$
\frac{0.0825}{4}=0.020625,\quad \ln(1.020625)\approx0.020425,
$$
and hence,  
$$
t\approx\frac{0.69315}{4\times0.020425}\approx\frac{0.69315}{0.08170}\approx8.48\text{ years}\,.
$$

**(d) Continuously:**  
Using $$ A=P\, e^{rt} $$ with continuous compounding, we have  
$$
t=\frac{\ln2}{r}=\frac{0.69315}{0.0825}\approx8.40\text{ years}\,.
$$

---

**21. Half-Life of Sm-151**

The decay of Sm-151 is modeled by  
$$
\frac{dy}{dt}=-0.0077y\,.
$$
Its solution is  
$$
y(t)=y(0)e^{-0.0077t}\,.
$$
The half-life $$ T $$ is defined by $$ y(T)= \frac{1}{2}y(0) $$:
$$
\frac{1}{2}y(0)=y(0)e^{-0.0077T}\,.
$$
Cancel $$ y(0) $$ and take the natural logarithm:
$$
\ln\left(\frac{1}{2}\right)=-0.0077T\,,
$$
so that  
$$
T=\frac{\ln2}{0.0077}\,.
$$
Numerically,  
$$
T\approx\frac{0.69315}{0.0077}\approx90.04\text{ years}\,.
$$

---

**22. Decay Constant $$ k $$ for Np-240**

For an isotope with half-life $$ T_{1/2}=65 $$ minutes and decay modeled by  
$$
\frac{dy}{dt}=-ky\,,
$$
the solution is  
$$
y(t)=y(0)e^{-kt}\,.
$$
The half-life condition $$ y(65)=\frac{1}{2}y(0) $$ gives  
$$
\frac{1}{2}=e^{-65k}\,.
$$
Taking natural logarithms:
$$
\ln\left(\frac{1}{2}\right)=-65k\,,
$$
so  
$$
k=\frac{\ln2}{65}\,.
$$
Numerically,  
$$
k\approx\frac{0.69315}{65}\approx0.01066\text{ per minute}\,.
$$
**19. For $$ r=4.75\% $$:** **(a) Annually:** Approximately 14.95 years **(b) Monthly:** Approximately 14.62 years **(c) Quarterly:** Approximately 14.67 years **(d) Continuously:** Approximately 14.59 years --- **20. For $$ r=8.25\% $$:** **(a) Annually:** Approximately 8.74 years **(b) Monthly:** Approximately 8.42 years **(c) Quarterly:** Approximately 8.48 years **(d) Continuously:** Approximately 8.40 years --- **21. Half-Life of Sm-151:** Approximately 90.04 years --- **22. Decay Constant $$ k $$ for Np-240:** Approximately 0.01066 per minute

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