Deposit 3000 today in an account paying 10 If you will need

Deposit $3000 today in an account paying 10%. If you will need $8000, how long will you have to wait?

Solution

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

Where,

P = principal amount (the initial amount you borrow or deposit), in this case $3000
r = annual rate of interest (as a decimal), in this case 0.10
t = number of years the amount is deposited or borrowed for, in this case we don\'t know it
A = amount of money accumulated after n years, including interest, in this case $8000
n = number of times the interest is compounded per year, in this case not given i.e let it default as 12

This gives us 8,000=3,000(1+0.10/12)^12t

After simplification, 8/3=(1.008)^12t

By taking the logarithm of both sides,

Log8/3=Log1.008^12t

12t = Log (8)- Log (3) /Log (1.008)

( Log (8) = 0.9030 , Log (3) = 0.4771, Log (1.008) = 0.0034 )

12t = 0.9030-0.4771 / 0.0034

                                12t = 0.4359/0.0034

                                12t = 128.20
                                    t = 10.6838

This gives us 10.6838 months. we can acheive our goals in 10.6 months if compounded monthly @ 10%.