Question
if i invest 300 dollars a year, compunding monthly at an interest rate of 7.5% for 8 years, how long will it take for it to become 1 million
Answers
GPT-4o mini
To determine how long it would take for an investment of $300 per year, compounded monthly at an interest rate of 7.5%, to grow to $1 million, we'll need to use the future value of a series formula given by:
\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \( FV \) is the future value of the investment (in this case, $1,000,000),
- \( P \) is the annual contribution (in this case, $300),
- \( r \) is the monthly interest rate (annual rate / 12),
- \( n \) is the total number of contributions (number of years × number of contributions per year).
1. **Calculate the Monthly Interest Rate:**
\[
r = \frac{0.075}{12} \approx 0.00625
\]
2. **Setting Up the Equation:**
You want to solve for \( n \) when \( FV = 1,000,000 \), \( P = 300 \), and \( r = 0.00625 \):
\[
1,000,000 = 300 \times \frac{(1 + 0.00625)^n - 1}{0.00625}
\]
3. **Rearranging to Solve for n:**
Multiply both sides by \( 0.00625 \):
\[
1,000,000 \times 0.00625 = 300 \times ((1 + 0.00625)^n - 1)
\]
\[
6250 = 300 \times ((1 + 0.00625)^n - 1)
\]
\[
\frac{6250}{300} = (1 + 0.00625)^n - 1
\]
\[
20.8333 = (1 + 0.00625)^n - 1
\]
\[
21.8333 = (1 + 0.00625)^n
\]
4. **Taking Logarithms:**
To solve for \( n \):
\[
n = \frac{\log(21.8333)}{\log(1 + 0.00625)}
\]
Calculating those values:
\[
\log(21.8333) \approx 1.339
\]
\[
\log(1.00625) \approx 0.00271
\]
Then,
\[
n \approx \frac{1.339}{0.00271} \approx 494.87
\]
5. **Calculate in Years:**
Since \( n \) is in months, divide by 12:
\[
\text{Years} \approx \frac{494.87}{12} \approx 41.24
\]
Therefore, it would take approximately **41.2 years** of investing $300 per year at a 7.5% interest rate compounded monthly for the investment to grow to $1 million.
\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \( FV \) is the future value of the investment (in this case, $1,000,000),
- \( P \) is the annual contribution (in this case, $300),
- \( r \) is the monthly interest rate (annual rate / 12),
- \( n \) is the total number of contributions (number of years × number of contributions per year).
1. **Calculate the Monthly Interest Rate:**
\[
r = \frac{0.075}{12} \approx 0.00625
\]
2. **Setting Up the Equation:**
You want to solve for \( n \) when \( FV = 1,000,000 \), \( P = 300 \), and \( r = 0.00625 \):
\[
1,000,000 = 300 \times \frac{(1 + 0.00625)^n - 1}{0.00625}
\]
3. **Rearranging to Solve for n:**
Multiply both sides by \( 0.00625 \):
\[
1,000,000 \times 0.00625 = 300 \times ((1 + 0.00625)^n - 1)
\]
\[
6250 = 300 \times ((1 + 0.00625)^n - 1)
\]
\[
\frac{6250}{300} = (1 + 0.00625)^n - 1
\]
\[
20.8333 = (1 + 0.00625)^n - 1
\]
\[
21.8333 = (1 + 0.00625)^n
\]
4. **Taking Logarithms:**
To solve for \( n \):
\[
n = \frac{\log(21.8333)}{\log(1 + 0.00625)}
\]
Calculating those values:
\[
\log(21.8333) \approx 1.339
\]
\[
\log(1.00625) \approx 0.00271
\]
Then,
\[
n \approx \frac{1.339}{0.00271} \approx 494.87
\]
5. **Calculate in Years:**
Since \( n \) is in months, divide by 12:
\[
\text{Years} \approx \frac{494.87}{12} \approx 41.24
\]
Therefore, it would take approximately **41.2 years** of investing $300 per year at a 7.5% interest rate compounded monthly for the investment to grow to $1 million.