To calculate how long it will take for the account to grow from $10,000 to $12,500 with a 4.5% annual interest rate compounded monthly, we can use the formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial deposit).
- \(r\) 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.
Given:
- \(A = 12,500\)
- \(P = 10,000\)
- \(r = 0.045\)
- \(n = 12\)
We need to solve for \(t\):
\[ 12500 = 10000 \left(1 + \frac{0.045}{12}\right)^{12t} \]
First, simplify the fraction inside the parentheses:
\[ 1 + \frac{0.045}{12} = 1 + 0.00375 = 1.00375 \]
Now rewrite the equation:
\[ 12500 = 10000 (1.00375)^{12t} \]
Next, divide both sides by 10,000:
\[ 1.25 = (1.00375)^{12t} \]
Next, take the logarithm of both sides. Using the natural logarithm (ln) for simplicity:
\[ \ln(1.25) = \ln((1.00375)^{12t}) \]
Using the property of logarithms that allows us to bring the exponent in front:
\[ \ln(1.25) = 12t \cdot \ln(1.00375) \]
Now, solve for \(t\):
\[ t = \frac{\ln(1.25)}{12 \cdot \ln(1.00375)} \]
Now, calculate the values:
First, calculate \(\ln(1.25)\):
\[ \ln(1.25) \approx 0.22314 \]
Next, calculate \(\ln(1.00375)\):
\[ \ln(1.00375) \approx 0.003738 \]
Now substitute back into the equation for \(t\):
\[ t \approx \frac{0.22314}{12 \cdot 0.003738} \] \[ t \approx \frac{0.22314}{0.044856} \approx 4.973 \text{ years} \]
This rounds to approximately 5 years. Therefore, the answer is:
B) 5 years.
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