Answer
(a) $18.356 \text{ years}$
(b) $18.310 \text{ years}$
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 $r=6\%=0.06$
$\text{Compounded monthly} \to n = 12$
Since the investment is to be tripled $ \hspace{20pt} \therefore A = 3P$
$3P=P \left(1+\dfrac{0.06}{12} \right)^{12 t}$
$3P= P \left(1+\dfrac{1}{200} \right)^{12 t}$
Divide both sides by $P$:
$3 = \left(\dfrac{201}{200} \right)^{12 t}$
Take the natual logarithm of both sides:
$\ln 3 = \ln{\left(\dfrac{201}{200} \right)^{12 t}}$
$\ln 3 = 12t \ln{\left(\dfrac{201}{200} \right)}$
$12t = \dfrac{\ln 3}{\ln{\left(\dfrac{201}{200} \right)}}$
$t = \dfrac{\ln 3}{12 \ln{\left(\dfrac{201}{200} \right)}}$
$t \approx \boxed{18.356 \text{ years}}$
(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 $r=6\%=0.06$
Since the investment is to be doubled $ \hspace{20pt} \therefore A = 3P$
$3P = P e^{0.06t}$
Divide both sides by $P$:
$3= e^{0.06t}$
Take the natual logarithm of both sides:
$\ln 3 = \ln e^{0.06t}$
$\ln 3 = 0.06t \ln e $
$\because \ln e = 1$
$\therefore \ln 3 = 0.06t $
$t = \dfrac{\ln 3}{0.06}$
$t \approx \boxed{18.310 \text{ years}}$
