Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To determine the annual interest rate required for Wouter to achieve his goal of having R15,000 after 10 years with an initial investment of R8,000, compounded monthly, we should follow these steps:
1. Identify the known variables:
- Initial investment (Principal, [tex]\( P \)[/tex]): R8,000
- Final amount (Amount, [tex]\( A \)[/tex]): R15,000
- Time period ([tex]\( t \)[/tex]): 10 years
- Compounding frequency ([tex]\( n \)[/tex]): 12 times per year (monthly)
2. Recall the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Here, [tex]\( r \)[/tex] is the annual interest rate and we need to solve for it.
3. Rearrange the formula to solve for the annual interest rate [tex]\( r \)[/tex]:
[tex]\[ \frac{A}{P} = \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln\left(\frac{A}{P}\right) = \ln\left(\left(1 + \frac{r}{n}\right)^{nt}\right) \][/tex]
Simplifying the right-hand side:
[tex]\[ \ln\left(\frac{A}{P}\right) = nt \cdot \ln\left(1 + \frac{r}{n}\right) \][/tex]
Solving for [tex]\(\ln\left(1 + \frac{r}{n}\right)\)[/tex]:
[tex]\[ \ln\left(1 + \frac{r}{n}\right) = \frac{\ln\left(\frac{A}{P}\right)}{nt} \][/tex]
Exponentiating both sides to remove the natural logarithm:
[tex]\[ 1 + \frac{r}{n} = \exp\left(\frac{\ln\left(\frac{A}{P}\right)}{nt}\right) \][/tex]
Isolating [tex]\( \frac{r}{n} \)[/tex]:
[tex]\[ \frac{r}{n} = \exp\left(\frac{\ln\left(\frac{A}{P}\right)}{nt}\right) - 1 \][/tex]
Solving for [tex]\( r \)[/tex]:
[tex]\[ r = n \left(\exp\left(\frac{\ln\left(\frac{A}{P}\right)}{nt}\right) - 1\right) \][/tex]
4. Plug in the known values:
[tex]\[ A = 15,000, \quad P = 8,000, \quad t = 10, \quad n = 12 \][/tex]
[tex]\[ r = 12 \left( \exp\left(\frac{\ln\left(\frac{15,000}{8,000}\right)}{12 \times 10}\right) - 1 \right) \][/tex]
5. Simplifying the calculations:
[tex]\[ r = 12 \left( \exp\left(\frac{\ln(1.875)}{120}\right) - 1 \right) \][/tex]
6. Compute the result using a calculator:
[tex]\[ \exp\left(\frac{\ln(1.875)}{120}\right) \approx 1.005 \quad \text{(rounded)} \][/tex]
[tex]\[ \frac{r}{n} = 1.005 - 1 = 0.005 \][/tex]
[tex]\[ r \approx 12 \times 0.005 = 0.06 \][/tex]
Converting to a percentage:
[tex]\[ r \approx 6.3\% \][/tex]
Thus, Wouter will need an annual interest rate of approximately 6.3%, compounded monthly, to achieve his goal of R15,000 in 10 years from an initial investment of R8,000.
1. Identify the known variables:
- Initial investment (Principal, [tex]\( P \)[/tex]): R8,000
- Final amount (Amount, [tex]\( A \)[/tex]): R15,000
- Time period ([tex]\( t \)[/tex]): 10 years
- Compounding frequency ([tex]\( n \)[/tex]): 12 times per year (monthly)
2. Recall the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Here, [tex]\( r \)[/tex] is the annual interest rate and we need to solve for it.
3. Rearrange the formula to solve for the annual interest rate [tex]\( r \)[/tex]:
[tex]\[ \frac{A}{P} = \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln\left(\frac{A}{P}\right) = \ln\left(\left(1 + \frac{r}{n}\right)^{nt}\right) \][/tex]
Simplifying the right-hand side:
[tex]\[ \ln\left(\frac{A}{P}\right) = nt \cdot \ln\left(1 + \frac{r}{n}\right) \][/tex]
Solving for [tex]\(\ln\left(1 + \frac{r}{n}\right)\)[/tex]:
[tex]\[ \ln\left(1 + \frac{r}{n}\right) = \frac{\ln\left(\frac{A}{P}\right)}{nt} \][/tex]
Exponentiating both sides to remove the natural logarithm:
[tex]\[ 1 + \frac{r}{n} = \exp\left(\frac{\ln\left(\frac{A}{P}\right)}{nt}\right) \][/tex]
Isolating [tex]\( \frac{r}{n} \)[/tex]:
[tex]\[ \frac{r}{n} = \exp\left(\frac{\ln\left(\frac{A}{P}\right)}{nt}\right) - 1 \][/tex]
Solving for [tex]\( r \)[/tex]:
[tex]\[ r = n \left(\exp\left(\frac{\ln\left(\frac{A}{P}\right)}{nt}\right) - 1\right) \][/tex]
4. Plug in the known values:
[tex]\[ A = 15,000, \quad P = 8,000, \quad t = 10, \quad n = 12 \][/tex]
[tex]\[ r = 12 \left( \exp\left(\frac{\ln\left(\frac{15,000}{8,000}\right)}{12 \times 10}\right) - 1 \right) \][/tex]
5. Simplifying the calculations:
[tex]\[ r = 12 \left( \exp\left(\frac{\ln(1.875)}{120}\right) - 1 \right) \][/tex]
6. Compute the result using a calculator:
[tex]\[ \exp\left(\frac{\ln(1.875)}{120}\right) \approx 1.005 \quad \text{(rounded)} \][/tex]
[tex]\[ \frac{r}{n} = 1.005 - 1 = 0.005 \][/tex]
[tex]\[ r \approx 12 \times 0.005 = 0.06 \][/tex]
Converting to a percentage:
[tex]\[ r \approx 6.3\% \][/tex]
Thus, Wouter will need an annual interest rate of approximately 6.3%, compounded monthly, to achieve his goal of R15,000 in 10 years from an initial investment of R8,000.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
