Question

under continuous compounding, the amount of time t in years required for an investment to double is a function of the interest rate r according to the formula:
$t = \frac{\ln 2}{r}$
(a) if you invest $7000, how long will it take the investment to reach $14,000 if the interest rate is 3.4%? round to one decimal place.
(b) if you invest $7000, how long will it take the investment to reach $14,000 if the interest rate is 6%? round to one decimal place.
(c) using the doubling time found in part (b), how long would it take a $7000 investment to reach $28,000 if the interest rate is 6%? round to one decimal place.
part 1 of 3
if the interest rate is 3.4%, a $7000 investment will reach $14,000 in approximately 20.4 years.
part 2 of 3
if the interest rate is 6%, a $7000 investment will reach $14,000 in approximately 11.6 years.
part: 2 / 3
part 3 of 3
if the interest rate is 6%, a $7000 investment will reach $28,000 in approximately years.

Explanation:

Step1: Analyze the growth factor

To reach $28,000 from $7,000, the investment needs to double twice (since $7000 \times 2 = 14000$, $14000 \times 2 = 28000$).

Step2: Use the doubling time from part (b)

From part (b), the doubling time at 6% interest rate is 11.6 years. So for two doublings, we multiply the doubling time by 2.
$t = 11.6\times2 = 23.2$

Answer:

23.2