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 answer

To create an exponential equation from the given recursive formula, we can start by analyzing the equation you've provided:

The recursive equation is:

\[ f(x) \approx f(x-1) \cdot 1.025 \]

This indicates that the balance grows by a factor of \(1.025\) each year. The initial balance given is \(f(0) = 2310\).

We can express \(f(x)\) in terms of an exponential function. The recursive relationship can be interpreted as:

\[ f(x) = f(0) \cdot (1.025)^x \]

Substituting \(f(0) = 2310\), we get:

\[ f(x) = 2310 \cdot (1.025)^x \]

Now to express this in terms of \(A(t)\) (the amount of money in the account after \(t\) years), we can use the notation:

\[ A(t) = 2310 \cdot (1.025)^t \]

Now we can calculate the amount of money in the account after \(t = 2\) years:

\[ A(2) = 2310 \cdot (1.025)^2 \]

Calculating \( (1.025)^2 \):

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

Now substituting this back into the equation for \(A(2)\):

\[ A(2) = 2310 \cdot 1.050625 \approx 2431.44 \]

Rounding to the nearest cent:

\[ A(2) \approx 2431.44 \]

Therefore, the amount of money in the account after 2 years is approximately $2431.44.