Answer
(a) $8.693\text{ years}$
(b) $8.664 \text{ years}$
Work Step by Step
(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 $
$t = \dfrac{\ln 2}{0.08}$
$t \approx \boxed{8.664 \text{ years}}$
