Suppose that $\mathrm{P}$ is invested in a savings account in which interest, $\mathrm{k}$, is compounded continuously at $5 \%$ per year. The balance $P(t)$ after time $t$, in years, is $P(t)=P e^{k t}$.
a) What is the exponential growth function in terms of $P$ and 0.05 ?
\[
P(t)=P e^{0.05 t}
\]
b) If $\$ 6,000$ is invested what is the balance after 2 years?
(Round to the nearest cent.)

Step 1 ：Suppose that \(P\) is invested in a savings account in which interest, \(k\), is compounded continuously at 5% per year. The balance \(P(t)\) after time \(t\), in years, is given by the formula \(P(t)=P e^{k t}\).

Step 2 ：The exponential growth function in terms of \(P\) and 0.05 is \(P(t)=P e^{0.05 t}\).

Step 3 ：If $6000 is invested, we can find the balance after 2 years by substituting \(P=6000\), \(t=2\), and \(k=0.05\) into the equation.

Step 4 ：Calculating \(P(t)=6000 e^{0.05 \times 2}\) gives a balance of approximately 6631.03.

Step 5 ：\(\boxed{6631.03}\) is the balance after 2 years when $6000 is invested.