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Sagot :
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.
[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.
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