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 ?
\[
\mathrm{P}(\mathrm{t})=\mathrm{P} e^{0.04 \mathrm{t}}
\]
b) If $\$ 9,000$ is invested what is the balance after 5 years?
\[
\$ 10992.62
\]
(Round to the nearest cent.)
c) When will an investment of $\$ 9,000$ double itself?
$\square$ years
(Round to the nearest tenth of a year.)

Step 1 ：The exponential growth function in terms of P and 0.04 is given by: \(P(t) = P \times e^{0.04t}\)

Step 2 ：If $9,000 is invested, the balance after 5 years can be calculated by substituting P = 9000 and t = 5 into the exponential growth function: \(P(5) = 9000 \times e^{0.04 \times 5} = 9000 \times e^{0.2} = 9000 \times 1.22140275816 = $10992.62\)

Step 3 ：To find out when an investment of $9,000 will double itself, we need to solve the equation \(9000 \times e^{0.04t} = 2 \times 9000\) for t. Simplifying, we get \(e^{0.04t} = 2\), then \(0.04t = \ln(2)\), and finally \(t = \frac{\ln(2)}{0.04} = 17.3\) years

Step 4 ：\(\boxed{17.3}\) years is the time it will take for an investment of $9,000 to double itself.