## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

(a) $8.693\text{ years}$ (b) $8.664 \text{ years}$
(a) The compound interest formula is $A = P \left(1+\dfrac{r}{n} \right)^{n t}$ where $P:$ The principal amount $r:$ Annual interest rate $n:$ Number of compoundings per year $t:$ Number of years $A:$ Amount after $t$ years Here we have $r=8\%=0.08$ $\text{Compounded monthly} \to n = 12$ Since the investment is to be doubled $\hspace{20pt} \therefore A = 2P$ $2P=P \left(1+\dfrac{0.08}{12} \right)^{12 t}$ $2P= P \left(1+\dfrac{1}{150} \right)^{12 t}$ Divide both sides by $P$: $2 = \left(\dfrac{151}{150} \right)^{12 t}$ Take the natual logarithm of both sides: $\ln 2 = \ln{\left(\dfrac{151}{150} \right)^{12 t}}$ $\ln 2 = 12t \ln{\left(\dfrac{151}{150} \right)}$ $12t = \dfrac{\ln 2}{\ln{\left(\dfrac{151}{150} \right)}}$ $t = \dfrac{\ln 2}{12 \ln{\left(\dfrac{151}{150} \right)}}$ $t \approx \boxed{8.693 \text{ years}}$ (b) Continuous Compounding Formula $$A = P e^{r t}$$ where $P:$ The principal amount $r:$ Annual interest rate $t:$ Number of years $A:$ Amount after $t$ years Here we have $r=8\%=0.08$ Since the investment is to be doubled $\hspace{20pt} \therefore A = 2P$ $2P = P e^{0.08t}$ Divide both sides by $P$: $2 = e^{0.08t}$ Take the natual logarithm of both sides: $\ln 2 = \ln e^{0.08t}$ $\ln 2 = 0.08t \ln e$ $\because \ln e = 1$ $\therefore \ln 2 = 0.08t$ $t = \dfrac{\ln 2}{0.08}$ $t \approx \boxed{8.664 \text{ years}}$