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

Algebra ->  Exponential-and-logarithmic-functions -> 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      Log On


   



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) About Me  (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.