Suppose that $\mathrm{P}$ is invested in a savings account in which interest, $k$, is compounded continuously at $4 \%$ per year. The balance $P(t)$ after time $\mathrm{t}$, in years, is $\mathrm{P}(\mathrm{t})=\mathrm{P} e^{\mathrm{kt}}$.
a) What is the exponential growth function in terms of $P$ and 0.04 ?
\[
P(t)=\square
\]
Answer
\(\boxed{P(t)=P e^{0.04t}}\) is the final answer.
Step 1 ：Suppose that $P$ is invested in a savings account in which interest, $k$, is compounded continuously at $4 \%$ per year. The balance $P(t)$ after time $t$, in years, is $P(t)=P e^{kt}$.
Step 2 ：What is the exponential growth function in terms of $P$ and 0.04 ?
Step 3 ：Substitute $k$ with 0.04 in the equation $P(t)=P e^{kt}$.
Step 4 ：So, the exponential growth function in terms of $P$ and 0.04 is $P(t)=P e^{0.04t}$.
Step 5 ：\(\boxed{P(t)=P e^{0.04t}}\) is the final answer.
