To solve this problem, we'll use the compound interest formula, which is given by:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount of money in the account.
- [tex]\( P \)[/tex] is the initial principal (the amount of money initially placed in the account).
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal).
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year.
- [tex]\( t \)[/tex] is the time in years.
Given:
- Initial amount ([tex]\( P \)[/tex]) = [tex]$7000
- Final amount (\( A \)) = $[/tex]14,322
- Annual interest rate ([tex]\( r \)[/tex]) = 5% = 0.05 (as a decimal)
- Number of times interest is compounded per year ([tex]\( n \)[/tex]) = 2 (since it is compounded semiannually)
We need to find the time ([tex]\( t \)[/tex]) it will take for the account to grow to [tex]$14,322.
Start by plugging the known values into the compound interest formula:
\[ 14322 = 7000 \left(1 + \frac{0.05}{2}\right)^{2t} \]
First, simplify the expression inside the parentheses:
\[1 + \frac{0.05}{2} = 1 + 0.025 = 1.025\]
So the equation becomes:
\[ 14322 = 7000 \cdot (1.025)^{2t} \]
Next, divide both sides by 7000 to isolate the exponential term:
\[ \frac{14322}{7000} = (1.025)^{2t} \]
\[ 2.046 = (1.025)^{2t} \]
Now, to solve for \( t \), we'll take the natural logarithm of both sides. This helps us use the property of logarithms that allows us to bring the exponent in front:
\[ \ln(2.046) = \ln((1.025)^{2t}) \]
Using the property \(\ln(a^b) = b \ln(a)\), this becomes:
\[ \ln(2.046) = 2t \cdot \ln(1.025) \]
Next, solve for \( t \) by dividing both sides by \( 2 \ln(1.025) \):
\[ t = \frac{\ln(2.046)}{2 \cdot \ln(1.025)} \]
Calculating the natural logarithms and performing the division (without rounding intermediate steps):
\[ \ln(2.046) \approx 0.7168 \]
\[ \ln(1.025) \approx 0.0247 \]
\[ t = \frac{0.7168}{2 \cdot 0.0247} \]
\[ t = \frac{0.7168}{0.0494} \]
\[ t \approx 14.495968478616676 \]
Finally, rounding to the nearest hundredth:
\[ t \approx 14.50 \]
So, it will take approximately 14.50 years for the account to grow to $[/tex]14,322.
