Question

Suppose that P is invested in a savings account in which interest, k, is compounded continuously at 3% per year. The

balance P(t) after time t, in years, is P(t) = Pekt.
a) What is the exponential growth function in terms of P and 0.03
P(1) =

Answer 1

For a time period of t = 1 year, the balance becomes P(1) = Pe^0.03.

Step-by-step explanation:

The question concerns the calculation of the balance P(t) after time t, in years, for an initial investment P when the interest is compounded continuously at a rate k.

The given function is P(t) = Pe^kt.

To express the exponential growth function in terms of P and the rate 0.03, we must identify k with 0.03 since the problem states that the interest is compounded continuously at 3% per year.

Thus, the exponential growth function becomes: P(t) = Pe^0.03t.

For t = 1 year, the balance would be:

P(1) = Pe^0.03(1)

= Pe^0.03