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: 1 / 3
part 2 of 3
if the interest rate is 6%, a $7000 investment will reach $14,000 in approximately \\(\square\\) years.

Explanation:

Step1: Convert rate to decimal

Given \( r = 6\% = 0.06 \).

Step2: Apply doubling - time formula

The formula for doubling time under continuous compounding is \( t=\frac{\ln2}{r} \).
Substitute \( r = 0.06 \) into the formula: \( t=\frac{\ln2}{0.06} \).
We know that \( \ln2\approx0.6931 \), so \( t=\frac{0.6931}{0.06}\approx11.6 \).

Answer:

11.6