How much would you need to deposit in an account now in order to have $2,000.00 in the account in 1799 days? Assume the account earns

Answer 1

2000 = Pe^r(1799)

2000/P = e^1799r

ln(2000/P) = 1799r

ln2000 -lnP = 1799r

lnP = ln2000-1799r

lnP = about 7.6 -1799r

P = e^(7.6 -1799r)

if r = rate of daily interest

but if r = rate of yearly interest, then 1799/365.25 = 4.926

and

P = about e^(7.6 -4.9r)

if r were to equal APR = 10%

then

P = e^(7.6-4.9(.10))

= e^(7.11)

= $1,221

but revise the calculations if the interest isn't compounded continuously.

Instead use the formula

A = P(1+r/n)^nt where n=1 if compounded once a year and t = number of years using 1799/365.25

2000 = P(1+r)^1799/365.25 if compounded annually

solve for P

ln(2000/P) = (1+r)^4.9

ln2000 - lnP = (1+r)^4.9

lnP = ln2000 - (1+r)^4.9

p = 7.6/(1+r)^4.9