The amount of money invested in a certain account increases according to the following function, where $y_{0}$ is the initial amount of the investment, and $y$ is the amount present at time $t$ (in years).
\[
y=y_{0} e^{0.025 t}
\]
After how many years will the initial investment be doubled? Do not round any intermediate computations, and round your answer to the nearest tenth.
years

Step 1 ：Given the function \(y=y_{0} e^{0.025 t}\), where \(y_{0}\) is the initial amount of the investment, and \(y\) is the amount present at time \(t\) (in years).

Step 2 ：We are asked to find the time it takes for the initial investment to double. This means we need to solve for \(t\) in the equation \(y = 2y_0\).

Step 3 ：We substitute \(2y_0\) for \(y\) in the given equation and solve for \(t\).

Step 4 ：Let's assume \(y0 = 1\) and \(y = 2\).

Step 5 ：Solving the equation gives \(t = 27.72588722239781\).

Step 6 ：Rounding to the nearest tenth, we get \(t = 27.7\).

Step 7 ：Final Answer: The initial investment will be doubled after approximately \(\boxed{27.7}\) years.