To determine how long it will take for an investment of $500 to grow to $850 at an interest rate of 12% compounded monthly, we can use the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (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.
Here, we have:
- \( A = 850 \)
- \( P = 500 \)
- \( r = 0.12 \) (12% as a decimal)
- \( n = 12 \) (compounded monthly)
Plugging the values into the formula, we want to find \( t \):
\[ 850 = 500 \left(1 + \frac{0.12}{12}\right)^{12t} \]
Calculating \( \frac{0.12}{12} \):
\[ \frac{0.12}{12} = 0.01 \]
So the formula simplifies to:
\[ 850 = 500 \left(1 + 0.01\right)^{12t} \]
This can be further simplified to:
\[ 850 = 500 (1.01)^{12t} \]
Dividing both sides by 500:
\[ \frac{850}{500} = (1.01)^{12t} \]
Calculating \( \frac{850}{500} \):
\[ 1.7 = (1.01)^{12t} \]
To solve for \( t \), we take the natural logarithm of both sides:
\[ \ln(1.7) = \ln((1.01)^{12t}) \]
Using the power rule of logarithms:
\[ \ln(1.7) = 12t \cdot \ln(1.01) \]
Now, solving for \( t \):
\[ t = \frac{\ln(1.7)}{12 \cdot \ln(1.01)} \]
Calculating the logarithms:
- \( \ln(1.7) \approx 0.5333 \)
- \( \ln(1.01) \approx 0.00995033 \)
Now substituting those values:
\[ t = \frac{0.5333}{12 \cdot 0.00995033} \]
Calculating the denominator:
\[ 12 \cdot 0.00995033 \approx 0.119404 \]
Now dividing:
\[ t \approx \frac{0.5333}{0.119404} \approx 4.46 \]
Thus, it will take approximately 4.46 years for $500 to grow to $850 at an interest rate of 12% compounded monthly.
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