SOLUTION: P₀ is invested in a savings account where interest is compounded continuously at 3.1% per year.
a) Express P(t) in terms of P₀ and 0.031
b) Suppose that 1000$ is inve
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Question 251449: P₀ is invested in a savings account where interest is compounded continuously at 3.1% per year.
a) Express P(t) in terms of P₀ and 0.031
b) Suppose that 1000$ is invested, what is the balance after 1 yr? 2 yrs?
c) When will an investment of 1000$ double itself?
Helpful formula:
When an amount of money P₀ is invested at interest rate k, compounded continuously, interest is computed every instant and added to the original amount. The balance P(t), after t years, is given by the exponential growth model
P(t)=P₀e^kt
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.
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