Compounded Interest Daily

Question

How long will it take to build $1000.00 to at least $7500.00 at 9% compounded daily? How many years? How many days?

Answer 1

the underlying formula that you want to use here is

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

for this scenario that means taking

7500 = 1000(1 + 0.09/365)^(365t)

solving that algebraically for the value of t yields

t = ln(7.5) / (365 * ln(1 + 0.09/365))

t = 22.39

meaning that it will take 22.39 years to grow to at least $7500 in this scenario

Answer 2

The equation for compound interest is:

A(t) = A₀(1 + r/n)^nt where A(t) is the amount at time "t" where "t" is measured in years, A₀ is the initial amount (at time t = 0), "r" is the annual (yearly) interest rate (written as a decimal), and "n" is the number of compounding periods.

In this case A₀ = 1000, r = 0.09 (although they don't identify this as annual interest rate, we can assume it is), A(t) = 7500, and n = 365 (the number of days in a year - note that despite the fact that the average number of days in a year is actually 365.25 because of leap year, banks use 365).

So:

A(t) = A₀(1 + r/n)^nt

7500 = 1000(1 + 0.09/365)^365t

7500/1000 = (1.000246575)^365t

log(7.5) = log(1.000246575)^365t

log(7.5) = 365t•log(1.000246575)

log(7.5)/log(1.000246575) = 365t

8172.569898 = 365t

t = 8172.569898/365

t = 22.39 years

This value actually results in a value slightly less than 7500 (results in $7499.61) and since they want "at least" $7500, we can round up to t = 22.4 years.

You would leave the answer as years. They are not asking for the number of days, they just want the interest compounded daily.

Answer 3

1000 = 7500(1+.09/365)^(365t)

solve for t, = 22.39 years.