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 the recursion f(x) = f(x−1)·1.032 with f(0)=10000, an exponential model is

A(t) = 10000·(1.032)^t.

Writing this with continuous exponent form (and using ln(1.032) ≈ 0.0315 ≈ 0.032 to two significant digits) gives

A(t) = 10000·e^{0.0315 t} ≈ 10000·e^{0.032 t}.

To find when A(t) > 11,000 solve
10000·e^{0.0315 t} > 11000 ⇒ e^{0.0315 t} > 1.1 ⇒ t > ln(1.1)/0.0315 ≈ 0.09531/0.0315 ≈ 3.03 years.

So the balance exceeds $11,000 after a little over 3 years. If you require a whole number of yearly account statements (balances given once per year, rounded to the nearest cent), the first year with more than $11,000 is year 4 (i.e., in 2004).