We are asked to find the amount of money which must be invested to have $4000 in an account at the end of five years if the annual interest rate is 8% compounded monthly.

We will use the following formula:

=>A(t) = P( 1 + r/n) ^nt

P = principal

A = the amount after t years

r = the investment rate

n= number of times interest compounds per year

t = number of years

=> A(t) = P( 1 + r/n) ^nt

=> 4000 = P ( 1 + .08/12)^12(5)

=> 4000 = 1.4898P

=> 2684.92 = P

**The amount to invest is 2684.92.**

The interest rate earned in the account is an annual rate of 8% compounded monthly.

So the monthly rate of interest is 8/12 = 2/3 %. Let the amount to be invested initially be P. In 5 years or 5*12 = 60 months this has to become $4000.

P*(1 + (2/3)%)^60 = 4000

=> P( 1+2/300)^60 = 4000

=> P(302/300)^60 = 4000

It is easier to use log here to determine P

log [ P(302/300)^60] = log 4000

=> log P + 60*log (302/300) = log 4000

=> log P = log 4000 - 60*log (302/300)

=> log P = 3.428

P = 10^3.428

=> P = 2684.84

**The initial amount to be deposited is $2684.84**

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