You're absolutely on the right track! The problem asks how long it will take for Dave’s $8,000 to grow to $12,000 at 3.7% annual interest compounded monthly.
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✔ Step 1: Use the compound interest formula
The formula is:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A = 12,000 \) (future value)
- \( P = 8,000 \) (principal)
- \( r = 0.037 \) (annual interest rate as a decimal)
- \( n = 12 \) (compounded monthly)
- \( t = ? \) (time in years — what we’re solving for)
---
✔ Step 2: Plug in the values
\[
12,000 = 8,000 \left(1 + \frac{0.037}{12}\right)^{12t}
\]
Divide both sides by 8,000:
\[
\frac{12,000}{8,000} = \left(1 + \frac{0.037}{12}\right)^{12t}
\]
\[
1.5 = \left(1 + 0.0030833...\right)^{12t}
\]
\[
1.5 = (1.0030833...)^{12t}
\]
---
✔ Step 3: Take the natural logarithm of both sides
\[
\ln(1.5) = \ln\left((1.0030833...)^{12t}\right)
\]
Use log power rule: \( \ln(a^b) = b \cdot \ln(a) \)
\[
\ln(1.5) = 12t \cdot \ln(1.0030833...)
\]
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✔ Step 4: Solve for \( t \)
First, calculate the logs:
- \( \ln(1.5) \approx 0.405465 \)
- \( \ln(1.0030833...) \approx \ln(1 + 0.037/12) \approx 0.003079 \) (using calculator)
Now plug in:
\[
0.405465 = 12t \cdot 0.003079
\]
\[
t = \frac{0.405465}{12 \cdot 0.003079} = \frac{0.405465}{0.036948} \approx 10.97
\]
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✔ Final Answer:
\[
\boxed{11} \text{ years (approximately)}
\]
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📝 Note about your work:
You got
t = 11, which is correct! 👏
Your steps are mostly accurate — you correctly set up the equation and took the natural log. The only small issue is that you wrote “Qn(1.5)” instead of “ln(1.5)”, but that seems like a handwriting error. Also, your final answer box says “11 = t”, which should be “t = 11” — but the math is solid!
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✔ So yes —
it will take approximately 11 years for Dave’s $8,000 to grow to $12,000 at 3.7% compounded monthly.
Great job!
Parent Tip: Review the logic above to help your child master the concept of compound interest practice worksheet.
