Say you have invested \( \$ 35000 \) at a yearly interest rate of \( 5 \% \) compounded weekly. How many years will it take for our investment to double in size? Years \( = \)
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To determine how many years it will take for your investment to double at a yearly interest rate of 5% compounded weekly, you can use the formula for compound interest: \[ A = P(1 + \frac{r}{n})^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial investment). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the number of years the money is invested for. In your case, \( P = 35000 \), \( r = 0.05 \), and \( A = 70000 \) (double the initial investment) with \( n = 52 \) (weekly compounding). Setting this up, we have: \[ 70000 = 35000(1 + \frac{0.05}{52})^{52t} \] Simplifying gives us: \[ 2 = (1 + \frac{0.05}{52})^{52t} \] Now, taking the natural logarithm of both sides: \[ \ln(2) = 52t \cdot \ln(1 + \frac{0.05}{52}) \] Now, solve for \( t \): \[ t = \frac{\ln(2)}{52 \cdot \ln(1 + \frac{0.05}{52})} \] Calculating this gives you approximately \( t \approx 13.86 \) years. So, it will take about 13.86 years for your investment to double!