Question
if i invest 300 dollars a month, 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 calculate how long it will take for an investment of $300 per month, compounded monthly at an interest rate of 7.5%, to grow to $1,000,000, we can use the future value of a series formula:
\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \(FV\) is the future value (in this case, $1,000,000).
- \(P\) is the monthly investment ($300).
- \(r\) is the monthly interest rate (annual rate divided by 12).
- \(n\) is the total number of contributions (months).
First, we need to convert the annual interest rate to a monthly rate:
\[
r = \frac{7.5\%}{12} = \frac{0.075}{12} \approx 0.00625
\]
Now we can set up the equation with \(FV\) = $1,000,000 and \(P\) = $300:
\[
1,000,000 = 300 \times \frac{(1 + 0.00625)^n - 1}{0.00625}
\]
Now we want to solve for \(n\). We can first isolate the term involving \(n\):
\[
\frac{1,000,000 \times 0.00625}{300} = (1 + 0.00625)^n - 1
\]
Calculating the left side:
\[
\frac{1,000,000 \times 0.00625}{300} = \frac{6250}{300} \approx 20.8333
\]
So, we have:
\[
20.8333 = (1 + 0.00625)^n - 1
\]
Adding 1 to both sides:
\[
21.8333 = (1.00625)^n
\]
Next, we take the natural logarithm (ln) of both sides to solve for \(n\):
\[
\ln(21.8333) = n \cdot \ln(1.00625)
\]
Now, calculating the left side:
\[
\ln(21.8333) \approx 3.075
\]
And the right side:
\[
\ln(1.00625) \approx 0.00623
\]
Now, we can solve for \(n\):
\[
n = \frac{3.075}{0.00623} \approx 493.3
\]
Finally, since \(n\) is in months, we convert months into years:
\[
\text{Years} = \frac{493.3}{12} \approx 41.1
\]
So, it will take approximately **41 years** for an investment of $300 per month, compounded monthly at an interest rate of 7.5%, to grow to $1,000,000.
\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]
Where:
- \(FV\) is the future value (in this case, $1,000,000).
- \(P\) is the monthly investment ($300).
- \(r\) is the monthly interest rate (annual rate divided by 12).
- \(n\) is the total number of contributions (months).
First, we need to convert the annual interest rate to a monthly rate:
\[
r = \frac{7.5\%}{12} = \frac{0.075}{12} \approx 0.00625
\]
Now we can set up the equation with \(FV\) = $1,000,000 and \(P\) = $300:
\[
1,000,000 = 300 \times \frac{(1 + 0.00625)^n - 1}{0.00625}
\]
Now we want to solve for \(n\). We can first isolate the term involving \(n\):
\[
\frac{1,000,000 \times 0.00625}{300} = (1 + 0.00625)^n - 1
\]
Calculating the left side:
\[
\frac{1,000,000 \times 0.00625}{300} = \frac{6250}{300} \approx 20.8333
\]
So, we have:
\[
20.8333 = (1 + 0.00625)^n - 1
\]
Adding 1 to both sides:
\[
21.8333 = (1.00625)^n
\]
Next, we take the natural logarithm (ln) of both sides to solve for \(n\):
\[
\ln(21.8333) = n \cdot \ln(1.00625)
\]
Now, calculating the left side:
\[
\ln(21.8333) \approx 3.075
\]
And the right side:
\[
\ln(1.00625) \approx 0.00623
\]
Now, we can solve for \(n\):
\[
n = \frac{3.075}{0.00623} \approx 493.3
\]
Finally, since \(n\) is in months, we convert months into years:
\[
\text{Years} = \frac{493.3}{12} \approx 41.1
\]
So, it will take approximately **41 years** for an investment of $300 per month, compounded monthly at an interest rate of 7.5%, to grow to $1,000,000.