Question

A high-yield savings account that compounds interest continuously was opened in 2000. The recursive equation
f
(
x
)
≈
f
(
x
−
1
)
⋅
1.032
shows the amount of money in the account, rounded to the nearest cent, for each year after 2000 where
f
(
0
)
=
10
,
000
. Use the recursive formula to create an exponential equation that models the account’s balance,
A
, after t years, rounding values to two significant digits. Using your exponential growth model, how many years did it take for the account to have more than $11,000.00?

Answer 1

From f(x) ≈ f(x − 1)·1.032 with f(0) = 10000 we get the exponential model (discrete annual growth)
A(t) = 10000·(1.032)^t.
The continuous-compounding equivalent is
A(t) = 10000·e^{t ln(1.032)} ≈ 10000·e^{0.031 t}
(where ln(1.032) ≈ 0.031, rounded to two significant digits).

Solve 10000·e^{0.031 t} > 11000:
t > ln(1.1)/0.031 ≈ 0.09531/0.031 ≈ 3.03 years.

So the balance exceeds $11,000 after about 3.03 years (a little after the start of 2003). If you check only at whole-year points using the recursive factor, f(3) ≈ $10,991.05 and f(4) ≈ $11,342.93, so it first exceeds $11,000 at year 4.