Answer :
To determine how long it will take for an initial investment of [tex]$6,000 to double given different annual interest rates, we need to use the formula for compound interest.
### Annual Compounding
The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
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 (as a decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested or borrowed for.
Since we want to double our money, \( A \) is \( 2P \) or $[/tex]12,000 in this case. Since we are considering annual compounding, [tex]\( n = 1 \)[/tex].
To solve for [tex]\( t \)[/tex], we rearrange the formula to get:
[tex]\[ 2P = P \left(1 + r\right)^{t} \][/tex]
Simplifying:
[tex]\[ 2 = \left(1 + r\right)^{t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln(2) = t \cdot \ln(1 + r) \][/tex]
[tex]\[ t = \frac{\ln(2)}{\ln(1 + r)} \][/tex]
Let’s compute the time [tex]\( t \)[/tex] for each interest rate:
(a) For an annual interest rate of 6% (0.06):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.06)} \approx 11.896 \, \text{years} \][/tex]
(b) For an annual interest rate of 10% (0.10):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.10)} \approx 7.273 \, \text{years} \][/tex]
(c) For an annual interest rate of 20% (0.20):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.20)} \approx 3.802 \, \text{years} \][/tex]
(d) For an annual interest rate of 30% (0.30):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.30)} \approx 2.642 \, \text{years} \][/tex]
### Semi-Annual Compounding
Now, if the compounding is done semi-annually, [tex]\( n = 2 \)[/tex]. The formula becomes:
[tex]\[ 2 = \left(1 + \frac{r}{2}\right)^{2t} \][/tex]
Taking the natural logarithm of both sides, we get:
[tex]\[ \ln(2) = 2t \cdot \ln\left(1 + \frac{r}{2}\right) \][/tex]
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{r}{2}\right)} \][/tex]
Let's compute the time [tex]\( t \)[/tex] for semi-annual compounding for each interest rate:
(a) For an annual interest rate of 6% (0.06):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.06}{2}\right)} \approx 46.900 \, \text{half-years} \][/tex]
Since we want the number of years, we multiply by 0.5:
[tex]\[ t \approx 23.450 \, \text{years} \][/tex]
(b) For an annual interest rate of 10% (0.10):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.10}{2}\right)} \approx 28.413 \, \text{half-years} \][/tex]
[tex]\[ t \approx 14.206 \, \text{years} \][/tex]
(c) For an annual interest rate of 20% (0.20):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.20}{2}\right)} \approx 14.545 \, \text{half-years} \][/tex]
[tex]\[ t \approx 7.273 \, \text{years} \][/tex]
(d) For an annual interest rate of 30% (0.30):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.30}{2}\right)} \approx 9.919 \, \text{half-years} \][/tex]
[tex]\[ t \approx 4.960 \, \text{years} \][/tex]
### Summary
Here are the results:
#### Annual Compounding
- 6% interest rate: [tex]\( \approx 11.896 \, \text{years} \)[/tex]
- 10% interest rate: [tex]\( \approx 7.273 \, \text{years} \)[/tex]
- 20% interest rate: [tex]\( \approx 3.802 \, \text{years} \)[/tex]
- 30% interest rate: [tex]\( \approx 2.642 \, \text{years} \)[/tex]
#### Semi-Annual Compounding
- 6% interest rate: [tex]\( \approx 23.450 \, \text{years} \)[/tex]
- 10% interest rate: [tex]\( \approx 14.206 \, \text{years} \)[/tex]
- 20% interest rate: [tex]\( \approx 7.273 \, \text{years} \)[/tex]
- 30% interest rate: [tex]\( \approx 4.960 \, \text{years} \)[/tex]
As can be seen, the time required to double the investment decreases when the interest rate increases. Additionally, semi-annual compounding generally results in shorter required times compared to annual compounding.
To solve for [tex]\( t \)[/tex], we rearrange the formula to get:
[tex]\[ 2P = P \left(1 + r\right)^{t} \][/tex]
Simplifying:
[tex]\[ 2 = \left(1 + r\right)^{t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln(2) = t \cdot \ln(1 + r) \][/tex]
[tex]\[ t = \frac{\ln(2)}{\ln(1 + r)} \][/tex]
Let’s compute the time [tex]\( t \)[/tex] for each interest rate:
(a) For an annual interest rate of 6% (0.06):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.06)} \approx 11.896 \, \text{years} \][/tex]
(b) For an annual interest rate of 10% (0.10):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.10)} \approx 7.273 \, \text{years} \][/tex]
(c) For an annual interest rate of 20% (0.20):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.20)} \approx 3.802 \, \text{years} \][/tex]
(d) For an annual interest rate of 30% (0.30):
[tex]\[ t = \frac{\ln(2)}{\ln(1 + 0.30)} \approx 2.642 \, \text{years} \][/tex]
### Semi-Annual Compounding
Now, if the compounding is done semi-annually, [tex]\( n = 2 \)[/tex]. The formula becomes:
[tex]\[ 2 = \left(1 + \frac{r}{2}\right)^{2t} \][/tex]
Taking the natural logarithm of both sides, we get:
[tex]\[ \ln(2) = 2t \cdot \ln\left(1 + \frac{r}{2}\right) \][/tex]
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{r}{2}\right)} \][/tex]
Let's compute the time [tex]\( t \)[/tex] for semi-annual compounding for each interest rate:
(a) For an annual interest rate of 6% (0.06):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.06}{2}\right)} \approx 46.900 \, \text{half-years} \][/tex]
Since we want the number of years, we multiply by 0.5:
[tex]\[ t \approx 23.450 \, \text{years} \][/tex]
(b) For an annual interest rate of 10% (0.10):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.10}{2}\right)} \approx 28.413 \, \text{half-years} \][/tex]
[tex]\[ t \approx 14.206 \, \text{years} \][/tex]
(c) For an annual interest rate of 20% (0.20):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.20}{2}\right)} \approx 14.545 \, \text{half-years} \][/tex]
[tex]\[ t \approx 7.273 \, \text{years} \][/tex]
(d) For an annual interest rate of 30% (0.30):
[tex]\[ t = \frac{\ln(2)}{2 \cdot \ln\left(1 + \frac{0.30}{2}\right)} \approx 9.919 \, \text{half-years} \][/tex]
[tex]\[ t \approx 4.960 \, \text{years} \][/tex]
### Summary
Here are the results:
#### Annual Compounding
- 6% interest rate: [tex]\( \approx 11.896 \, \text{years} \)[/tex]
- 10% interest rate: [tex]\( \approx 7.273 \, \text{years} \)[/tex]
- 20% interest rate: [tex]\( \approx 3.802 \, \text{years} \)[/tex]
- 30% interest rate: [tex]\( \approx 2.642 \, \text{years} \)[/tex]
#### Semi-Annual Compounding
- 6% interest rate: [tex]\( \approx 23.450 \, \text{years} \)[/tex]
- 10% interest rate: [tex]\( \approx 14.206 \, \text{years} \)[/tex]
- 20% interest rate: [tex]\( \approx 7.273 \, \text{years} \)[/tex]
- 30% interest rate: [tex]\( \approx 4.960 \, \text{years} \)[/tex]
As can be seen, the time required to double the investment decreases when the interest rate increases. Additionally, semi-annual compounding generally results in shorter required times compared to annual compounding.
