To model the financial growth with an exponential equation, we first need to represent the growth of the account over time as \( y = a \cdot b^x \), where:
- \( y \) is the amount in the account after \( x \) years,
- \( a \) is the initial amount (or the amount at \( x = 0 \)),
- \( b \) is the growth factor per year,
- \( x \) is the number of years.
We know:
- After 6 years, \( y = 12430.42 \).
- After 7 years, \( y = 13156.99 \).
- Calculate the annual growth factor: \[ \text{Growth factor} = \frac{\text{Amount after 7 years}}{\text{Amount after 6 years}} = \frac{13156.99}{12430.42} \approx 1.0585 \]
This means that each year, the account grows by a factor of approximately \( 1.0585 \).
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Find \( a \): Since we don't have the initial amount (at \( x = 0 \)), we can use the value at \( x = 6 \): \[ 12430.42 = a \cdot (1.0585)^6 \] First, calculate \( (1.0585)^6 \): \[ (1.0585)^6 \approx 1.3743 \] Now, solve for \( a \): \[ 12430.42 = a \cdot 1.3743 \] \[ a \approx \frac{12430.42}{1.3743} \approx 9050.00 \]
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Write the exponential equation: Now, substituting back for \( a \) and \( b \) in the equation: \[ y = 9050 \cdot (1.0585)^x \]
So, the exponential growth equation that models this savings account is: \[ y = 9050 \cdot 1.06^{x} \]
Final Answer:
- \( y = 9050 \cdot (1.06)^x \) (rounded \( b \) value assuming rounding \( 1.0585 \) to two decimal places as \( 1.06 \)).