How long do you need to invest your money in an account earning an annual interest rate of \( 3.341 \% \) dampounded daily so that your investment grows from \( \$ 1,067.51 \) to \( \$ 10,553.00 \) over that period of time? Note: Your answer should be an integer)

We start with the formula for compound interest with daily compounding: \[ A = P\left(1+\frac{r}{365}\right)^{365t} \] Here, - \( P = 1067.51 \) is the initial investment, - \( A = 10553.00 \) is the final amount, - \( r = 0.03341 \) is the annual interest rate expressed as a decimal, - \( t \) is the number of years. Substitute the values into the formula: \[ 10553.00 = 1067.51\left(1+\frac{0.03341}{365}\right)^{365t} \] Divide both sides by \( 1067.51 \): \[ \left(1+\frac{0.03341}{365}\right)^{365t} = \frac{10553.00}{1067.51} \] Take the natural logarithm of both sides: \[ 365t \ln\left(1+\frac{0.03341}{365}\right) = \ln\left(\frac{10553.00}{1067.51}\right) \] Solve for \( t \): \[ t = \frac{\ln\left(\frac{10553.00}{1067.51}\right)}{365\,\ln\left(1+\frac{0.03341}{365}\right)} \] Notice that for small values, the approximation \( \ln\left(1 + \frac{0.03341}{365}\right) \approx \frac{0.03341}{365} \) is valid. Substituting this approximation, we have: \[ t \approx \frac{\ln\left(\frac{10553.00}{1067.51}\right)}{365\left(\frac{0.03341}{365}\right)} = \frac{\ln\left(\frac{10553.00}{1067.51}\right)}{0.03341} \] Now, calculate the ratio: \[ \frac{10553.00}{1067.51} \approx 9.887 \] Then, compute the natural logarithm: \[ \ln(9.887) \approx 2.2906 \] Thus, \[ t \approx \frac{2.2906}{0.03341} \approx 68.59 \] Since the answer should be given as an integer, we round to the nearest year: \[ t \approx 69 \] Therefore, you need to invest your money for \(\boxed{69}\) years.