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Sagot :
To determine how long it will take for [tex]$27,000 to grow to $[/tex]31,340.37[tex]$ when the interest rate is 5% compounded quarterly, we can use the compound interest formula:
\[ A = P \left(1 + \frac{r}{k}\right)^{kt} \]
Here:
- \( A \) is the final amount ($[/tex]31,340.37).
- [tex]\( P \)[/tex] is the principal amount ([tex]$27,000). - \( r \) is the annual interest rate (0.05). - \( k \) is the number of times interest is compounded per year (4, quarterly). - \( t \) is the number of years, which we need to find. First, we plug the known values into the formula: \[ 31,340.37 = 27,000 \left(1 + \frac{0.05}{4}\right)^{4t} \] Next, simplify the term inside the parenthesis: \[ 1 + \frac{0.05}{4} = 1 + 0.0125 = 1.0125 \] Now our equation is: \[ 31,340.37 = 27,000 \cdot 1.0125^{4t} \] To solve for \( t \), we first isolate the exponential part by dividing both sides by 27,000: \[ \frac{31,340.37}{27,000} = 1.0125^{4t} \] \[ 1.1614962963 = 1.0125^{4t} \] Take the natural logarithm (ln) on both sides to bring down the exponent: \[ \ln(1.1614962963) = \ln(1.0125^{4t}) \] Using the property of logarithms \( \ln(a^b) = b \cdot \ln(a) \): \[ \ln(1.1614962963) = 4t \cdot \ln(1.0125) \] Next, solve for \( t \): \[ t = \frac{\ln(1.1614962963)}{4 \cdot \ln(1.0125)} \] From the given answers, we know the following values: \[ \ln(1.1614962963) \approx 0.14907017685257912 \] \[ \ln(1.0125) \approx 0.01242251999855711 \] Thus, \[ t = \frac{0.14907017685257912}{4 \cdot 0.01242251999855711} \] \[ t = \frac{0.14907017685257912}{0.04969007999422844} \] \[ t \approx 2.9999987295229453 \] Therefore, approximately 3 years will be needed for the $[/tex]27,000 to grow to [tex]$31,340.37$[/tex] at an annual interest rate of 5% compounded quarterly.
- [tex]\( P \)[/tex] is the principal amount ([tex]$27,000). - \( r \) is the annual interest rate (0.05). - \( k \) is the number of times interest is compounded per year (4, quarterly). - \( t \) is the number of years, which we need to find. First, we plug the known values into the formula: \[ 31,340.37 = 27,000 \left(1 + \frac{0.05}{4}\right)^{4t} \] Next, simplify the term inside the parenthesis: \[ 1 + \frac{0.05}{4} = 1 + 0.0125 = 1.0125 \] Now our equation is: \[ 31,340.37 = 27,000 \cdot 1.0125^{4t} \] To solve for \( t \), we first isolate the exponential part by dividing both sides by 27,000: \[ \frac{31,340.37}{27,000} = 1.0125^{4t} \] \[ 1.1614962963 = 1.0125^{4t} \] Take the natural logarithm (ln) on both sides to bring down the exponent: \[ \ln(1.1614962963) = \ln(1.0125^{4t}) \] Using the property of logarithms \( \ln(a^b) = b \cdot \ln(a) \): \[ \ln(1.1614962963) = 4t \cdot \ln(1.0125) \] Next, solve for \( t \): \[ t = \frac{\ln(1.1614962963)}{4 \cdot \ln(1.0125)} \] From the given answers, we know the following values: \[ \ln(1.1614962963) \approx 0.14907017685257912 \] \[ \ln(1.0125) \approx 0.01242251999855711 \] Thus, \[ t = \frac{0.14907017685257912}{4 \cdot 0.01242251999855711} \] \[ t = \frac{0.14907017685257912}{0.04969007999422844} \] \[ t \approx 2.9999987295229453 \] Therefore, approximately 3 years will be needed for the $[/tex]27,000 to grow to [tex]$31,340.37$[/tex] at an annual interest rate of 5% compounded quarterly.
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