Question 251449
the formula for continuous compounding is:



f = p*e^(r*t)


f = future value
p = present amount
e = scientific constant whose value is equal to 2.718281828...
r = interest rate per year
t = number of years


note:


interest rate percent / 100% = interest rate.


in your problem, the interest rate per year would be:


3.1% / 100% = .031


p = $1,000.00


for a value after 1 year, t = 1
for a value after 2 years, t = 2
how to solve for when the investment doubles comes after that.


value of $1,000 investment after 1 year.


p = $1,000
r = .031
t = 1


f = p * e^(r*t) becomes:


f = 1000 * e^(.031*1)


solve for f to get:


f = 1031.485504


value of $1,000 investment after 2 years.


p = $1,000
r = .031
t = 2


f = p * e^(r*t) becomes:


f = 1000 * e^(.031*2)


solve for f to get:


f = 1063.962345


how long does it take for the money to double.


let p = 1
let f = 2


formula of f = p * e^(r*t) becomes:


2 = 1 * e^(.031*t)


you need to solve for t.


2 = 1 * e^(.031*t) becomes:


2 = e^(.031*t)


take the log of both sides of this equation to get:


log(2) = log(e^(.031*t))


since log (a^b) = b*log(a), your equation becomes:


log(2) = .031*t*log(e)


divide both sides of this equation by .031 * log(3) to get


t = log(2) / (.031*log(e))


solve for t to get:


t = 22.35958647


confirm by substituting for t in the original equation.


2 = 1 * e^(.031*t) becomes:


2 = 1 * e^(.031*22.35958647) which becomes:


2 = 1 * e^(.693147181) which becomes:


2 = 1 * 2 which becomes:


2 = 2 confirming that the solution is valid.


your answer is:


the investment will double in 22.35958647 years.