Let's tackle this step-by-step.
### Step 1: Determine the function for the value of the account after [tex]\( t \)[/tex] years
We are given:
- Initial investment ([tex]\(P\)[/tex]) = \[tex]$4500
- Annual interest rate (\(r\)) = 3% = 0.03
- Compounded annually (\(n\)) = 1
The formula for compound interest is:
\[ f(t) = P \left(1 + \frac{r}{n}\right)^{nt} \]
Plugging in the given values, we get:
\[ f(t) = 4500 \left(1 + \frac{0.03}{1}\right)^{1 \cdot t} \]
which simplifies to:
\[ f(t) = 4500 \left(1.03\right)^{t} \]
So, the function \( f(t) \) is:
\[ f(t) = 4500 \times 1.03^t \]
### Step 2: Calculate the value of the account after 11 years
Now, we need to find the value of the account after 11 years (\(t = 11\)):
\[ f(11) = 4500 \times 1.03^{11} \]
The value of the account after 11 years is:
\[ \approx \$[/tex]6229.05 \]
### Step 3: Find when the value of the account will be [tex]$6300
We need to find the time \(t\) when the value of the account is \$[/tex]6300. We have:
[tex]\[ 6300 = 4500 \times 1.03^t \][/tex]
To solve for [tex]\(t\)[/tex], we can rearrange the equation:
[tex]\[ 1.03^t = \frac{6300}{4500} \][/tex]
[tex]\[ 1.03^t = 1.4 \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ t \ln(1.03) = \ln(1.4) \][/tex]
Dividing both sides by [tex]\(\ln(1.03)\)[/tex]:
[tex]\[ t = \frac{\ln(1.4)}{\ln(1.03)} \][/tex]
The value of [tex]\(t\)[/tex] is:
[tex]\[ \approx 11.383 \text{ years} \][/tex]
### Summary
- The function for the value of the account after [tex]\(t\)[/tex] years is:
[tex]\[ f(t) = 4500 \times 1.03^t \][/tex]
- The value of the account after 11 years is approximately:
[tex]\[\$6229.05\][/tex]
- The account will reach \$6300 after approximately:
[tex]\[ t \approx 11.383 \text{ years} \][/tex]
