Suppose that $1400 is invested at an interest rate of 4% per year, compounded continuously. After how many years will the initial investment be doubled? Round your answer to two decimal places.
Use the compounded continuously interest formula a = Pe^(rt), where a is the future amount, P is the principal, r is the annual interest rate, and t is the time in years. If the investment is doubled, then a = 2*1400 = $2800. You are solving for t.
a = Pe^(rt)
2800 = 1400e^(0.04t) (divide both sides by 14000)
2 = e^(0.04t) (take natural log of both sides)
ln (2) = ln (e^(0.04t)) (simplify natural log)
ln (2) = 0.04t (divide both sides by 0.04)
ln (2) / 0.04 = t
t = 17.3286795
ANSWER: about 17.33 years
*If the investment wasn't compounded continuously, then you would have used the compounded interest formula a = P(1 + r/n)^(nt), where a is the future value, P is the principal, r is the annual interest rate, n is the number of times it is a compounded a year (like if it was compounded monthly, then n = 12), and t is the time in years.
Use the compounded continuously interest formula a = Pe^(rt), where a is the future amount, P is the principal, r is the annual interest rate, and t is the time in years. If the investment is doubled, then a = 2*1400 = $2800. You are solving for t.
a = Pe^(rt)
2800 = 1400e^(0.04t) (divide both sides by 14000)
2 = e^(0.04t) (take natural log of both sides)
ln (2) = ln (e^(0.04t)) (simplify natural log)
ln (2) = 0.04t (divide both sides by 0.04)
ln (2) / 0.04 = t
t = 17.3286795
ANSWER: about 17.33 years
*If the investment wasn't compounded continuously, then you would have used the compounded interest formula a = P(1 + r/n)^(nt), where a is the future value, P is the principal, r is the annual interest rate, n is the number of times it is a compounded a year (like if it was compounded monthly, then n = 12), and t is the time in years.