Question

A high-yield savings account that compounds interest continuously was opened in 2010. The recursive equation f(x)≈f(x−1)⋅1.025
shows the amount of money in the account, rounded to the nearest cent for each year after 2010, in which f(0)=2,310
. 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 much money is in the account after 2 years?(1 point)
$

Answer 1

To derive the exponential equation from the given recursive formula, we start with the information:

1. The recursive equation is:
\[
f(x) \approx f(x-1) \cdot 1.025
\]

2. The initial amount at \( t = 0 \), or \( f(0) \), is 2310.

From the recursive formula, we see that each year the value increases by a factor of 1.025. We can express this recursively as follows:

- For \( f(1) \):
\[
f(1) \approx f(0) \cdot 1.025 = 2310 \cdot 1.025
\]

- For \( f(2) \):
\[
f(2) \approx f(1) \cdot 1.025 \approx (2310 \cdot 1.025) \cdot 1.025 = 2310 \cdot (1.025^2)
\]

In general, we can express the balance after \( t \) years (where \( t \) corresponds to \( x \)) as:
\[
f(t) \approx 2310 \cdot (1.025^t)
\]

This creates an exponential growth model:
\[
A(t) = 2310 \cdot (1.025^t)
\]

Now, we want to calculate how much money is in the account after 2 years (\( t = 2 \)):
\[
A(2) = 2310 \cdot (1.025^2)
\]

Calculating \( 1.025^2 \):
\[
1.025^2 \approx 1.050625
\]

Therefore:
\[
A(2) \approx 2310 \cdot 1.050625 \approx 2432.44
\]

Finally, rounding to two significant digits, we find:
\[
A(2) \approx 2430.00
\]

So, the amount of money in the account after 2 years is approximately:
\[
\boxed{2430}
\]