To determine how long it will take for an investment to grow from $500 to $850 at an interest rate of 12% compounded monthly, we can use the future value formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \(A\) is the future value of the investment ($850),
- \(P\) is the principal amount ($500),
- \(r\) is the annual interest rate (12% or 0.12),
- \(n\) is the number of times interest is compounded per year (monthly, so \(n = 12\)),
- \(t\) is the number of years.
Now, we can rearrange the formula to solve for \(t\):
\[ 850 = 500 \left(1 + \frac{0.12}{12}\right)^{12t} \]
First, calculate \(\frac{0.12}{12}\):
\[ \frac{0.12}{12} = 0.01 \]
Now we can substitute back:
\[ 850 = 500 \left(1 + 0.01\right)^{12t} \] \[ 850 = 500 (1.01)^{12t} \]
Next, divide both sides by 500:
\[ \frac{850}{500} = (1.01)^{12t} \] \[ 1.7 = (1.01)^{12t} \]
Now, take the natural logarithm of both sides:
\[ \ln(1.7) = \ln\left((1.01)^{12t}\right) \] \[ \ln(1.7) = 12t \cdot \ln(1.01) \]
Now, solve for \(t\):
\[ t = \frac{\ln(1.7)}{12 \cdot \ln(1.01)} \]
Now we can calculate the values:
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Calculate \(\ln(1.7)\): \(\ln(1.7) \approx 0.5316\)
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Calculate \(\ln(1.01)\): \(\ln(1.01) \approx 0.00995\)
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Plug these values in:
\[ t \approx \frac{0.5316}{12 \cdot 0.00995} \approx \frac{0.5316}{0.1194} \approx 4.44 \]
Thus, it will take approximately 4.44 years for $500 to accumulate to $850 at 12% interest compounded monthly.