Question

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).
$y = y_0 e^{0.0475t}$
after how many years will the initial investment be doubled? do not round any intermediate computations, and round your answer to the nearest tenth.
\boxed{} years

Explanation:

Step1: Set y = 2y₀

We want to find when the investment doubles, so set \( y = 2y_0 \) in the equation \( y = y_0 e^{0.0475t} \). This gives \( 2y_0 = y_0 e^{0.0475t} \).

Step2: Divide by y₀

Divide both sides by \( y_0 \) (assuming \( y_0
eq 0 \)), we get \( 2 = e^{0.0475t} \).

Step3: Take natural log

Take the natural logarithm of both sides: \( \ln(2) = \ln(e^{0.0475t}) \). Using the property \( \ln(e^x)=x \), this simplifies to \( \ln(2) = 0.0475t \).

Step4: Solve for t

Solve for \( t \) by dividing both sides by 0.0475: \( t = \frac{\ln(2)}{0.0475} \). Calculate \( \ln(2) \approx 0.6931 \), so \( t \approx \frac{0.6931}{0.0475} \approx 14.6 \).

Answer:

14.6