# Noha is investing $2517 in an account compounded monthly. She wants to have $3000 in 3 years for a trip. What interest rate, compounded monthly,does she need?To the nearest hundredth of a precent

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### 2 Answers

Noha is investing $2517 in an account. Let the rate at which the amount is compounded monthly so that it is equal to $3000 after 3 years be r.

If the initial amount is P and the rate of interest is r, the amount after t billing cycles is equal to `P*(1 + r)^t` . Substituting the values given in the question:

`3000 = 2517*(1 + r)^36`

=> `(1 +r)^36 = 3000/2517`

Take the log of both the sides

=> `log(1 + r) = (log(3000/2517))/36`

=> `log(1 +r) ~~ 2.1177*10^-3`

Converting the logarithm to exponential function

`1 + r ~~ 1.004888`

=> `r ~~ 0.00488`

Converting this to an annualized rate of interest gives `0.00488*12*100 ~~ 5.87%`

**The annual rate of interest required is 5.87%**

Begin with the compound interest formula, *A* = *P*(1 + *r*/*n*)^*nt*, where *A* is the amount in the account after some time period of investment, *P* is the amount invested, *r* is the interest rate, *n* is the number of times per year interest is earned, and *t* is the length of time of the investment. In this problem, *A* = $3000, *P* = $2517, *n* = 12, and *t* = 3, which leaves *r* for us to find. Substitute in the known values and solve for *r*:

3000 = 2517(1 + *r*/12)^(12*3)

3000/2517 = (1 + *r*/12)^36

(3000/2517)^(1/36) = 1 + *r*/12

(3000/2517)^(1/36) - 1 = *r*/12

[(3000/2517)^(1/36) - 1]*12 = *r *

**To the nearest hundredth of a percent, we find r to be approximately 5.87%.**