Problem

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 ? \[ P(t)=\square \]

Solution

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.

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Source: https://solvelyapp.com/problems/7yMUzuHXfM/

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