To determine the number of years it will take for the initial investment to double, we can use the given function:
[tex]\[ y = y_0 e^{0.055t} \][/tex]
Here, [tex]\( y_0 \)[/tex] represents the initial investment, and [tex]\( y \)[/tex] represents the amount present at time [tex]\( t \)[/tex] years. We want to find the time [tex]\( t \)[/tex] when the initial investment [tex]\( y_0 \)[/tex] is doubled, which means:
[tex]\[ y = 2y_0 \][/tex]
To proceed, we substitute [tex]\( y = 2y_0 \)[/tex] into the given exponential equation:
[tex]\[ 2y_0 = y_0 e^{0.055t} \][/tex]
Next, we can simplify this equation by dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[ 2 = e^{0.055t} \][/tex]
To solve for [tex]\( t \)[/tex], we need to take the natural logarithm (ln) on both sides of the equation:
[tex]\[ \ln(2) = \ln(e^{0.055t}) \][/tex]
Since the natural logarithm and the exponential function are inverses of each other, we can simplify the right-hand side:
[tex]\[ \ln(2) = 0.055t \][/tex]
Now, we solve for [tex]\( t \)[/tex] by dividing both sides by 0.055:
[tex]\[ t = \frac{\ln(2)}{0.055} \][/tex]
Using the known value of [tex]\( \ln(2) \approx 0.6931471805599453 \)[/tex]:
[tex]\[ t = \frac{0.6931471805599453}{0.055} \approx 12.602676010180824 \][/tex]
Finally, we round the result to the nearest tenth:
[tex]\[ t \approx 12.6 \][/tex]
Thus, the number of years required for the investment to double is approximately:
[tex]\[ \boxed{12.6} \][/tex]
