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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 \( = \)

Ask by Mccoy Alexander. in South Africa
Mar 08,2025

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Answer

It will take approximately 13.87 years for the investment to double in size.

Solution

To find out how many years it will take for the investment to double in size, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after \( n \) years, including interest. - \( P \) is the principal amount (initial investment). - \( r \) is the annual interest rate (in decimal form). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years. Given: - \( P = \$35000 \) - \( r = 5\% = 0.05 \) (in decimal form) - \( n = 52 \) (compounded weekly) We want to find \( t \) when \( A = 2P \). Substitute the values into the formula: \[ 2P = P \left(1 + \frac{0.05}{52}\right)^{52t} \] Simplify the equation: \[ 2 = \left(1 + \frac{0.05}{52}\right)^{52t} \] Now, we can solve for \( t \) using logarithms. Solve the equation by following steps: - step0: Solve for \(t\): \(2=\left(1+\frac{0.05}{52}\right)^{52t}\) - step1: Simplify: \(2=\left(\frac{1041}{1040}\right)^{52t}\) - step2: Swap the sides: \(\left(\frac{1041}{1040}\right)^{52t}=2\) - step3: Take the logarithm of both sides: \(\log_{\frac{1041}{1040}}{\left(\left(\frac{1041}{1040}\right)^{52t}\right)}=\log_{\frac{1041}{1040}}{\left(2\right)}\) - step4: Evaluate the logarithm: \(52t=\log_{\frac{1041}{1040}}{\left(2\right)}\) - step5: Divide both sides: \(\frac{52t}{52}=\frac{\log_{\frac{1041}{1040}}{\left(2\right)}}{52}\) - step6: Divide the numbers: \(t=\frac{\log_{\frac{1041}{1040}}{\left(2\right)}}{52}\) The solution to the equation is \( t = \frac{\log_{\frac{1041}{1040}}{(2)}}{52} \). Now, we can calculate the value of \( t \) to find out how many years it will take for the investment to double in size. Calculate the value by following steps: - step0: Calculate: \(\frac{\log_{\frac{1041}{1040}}{\left(2\right)}}{52}\) It will take approximately 13.87 years for the investment to double in size.

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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!

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