Answer :
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.
