Question

how much must you deposit in an account that pays 8% interest, compounded monthly, to have a balance of $1000 after 3 years?

Answer 1

1000=N(1 + .08/12)^36 where N is the initial deposit.

So 1000=N(1.00666667)^36

N=1000/(1.00666667)^36

ln N=ln 1000/(1.00666667)^36=ln 1000-36 ln (1.00666667)

ln N=6.6685517411100682158933244766027

N=e^6.6685517411100682158933244766027=$787.25463 as the initial deposit.