Suppose that $\$ 1100$ is invested at an interest rate of $3.25 \%$ per year, compounded continuously. After how many years will the initial investment be doubled?
Do not round any intermediate computations, and round your answer to the nearest hundredth.
years

Step 1 ：Given that the principal amount (P) is $1100, the final amount (A) is $2200 (since the investment is to be doubled), and the interest rate (r) is 0.0325 (or 3.25%).

Step 2 ：We are to find the time (t) in years it takes for the investment to double.

Step 3 ：We use the formula for continuous compounding, which is \(A = Pe^{rt}\), where A is the final amount, P is the principal amount, r is the interest rate, and t is the time in years.

Step 4 ：Since we want to find out when the initial investment will be doubled, we can substitute A = 2P into the formula and solve for t.

Step 5 ：Substituting the given values into the formula, we get \(2200 = 1100e^{0.0325t}\).

Step 6 ：Solving for t, we get \(t = \frac{\ln(\frac{2200}{1100})}{0.0325} \approx 21.327605555690624\).

Step 7 ：Rounding to the nearest hundredth, we get \(t \approx 21.33\) years.

Step 8 ：Final Answer: The initial investment will be doubled after approximately \(\boxed{21.33}\) years.