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Write an exponential function \( y=a b^{x} \) whose graph passes through \( (3,56) \) and \( (5,224) . \quad y=a b^{x} \)

Ask by Chen Brooks. in the United States
Mar 10,2025

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Answer

The exponential function is \( y = 7 \cdot 2^{x} \).

Solution

1. Since the graph of the function \( y = ab^{x} \) passes through the point \( (3,56) \), we have \[ 56 = a b^{3}. \] 2. Similarly, because it passes through \( (5,224) \), we have \[ 224 = a b^{5}. \] 3. To eliminate \( a \), we divide the second equation by the first: \[ \frac{224}{56} = \frac{a b^{5}}{a b^{3}} \quad \Rightarrow \quad 4 = b^{2}. \] 4. Solving for \( b \): \[ b^{2} = 4 \quad \Rightarrow \quad b = 2. \] (We take \( b = 2 \) since the base of an exponential function is typically positive.) 5. Substitute \( b = 2 \) back into the equation \( 56 = a b^{3} \): \[ 56 = a \cdot 2^{3} \quad \Rightarrow \quad 56 = 8a \quad \Rightarrow \quad a = 7. \] 6. Therefore, the exponential function is: \[ y = 7 \cdot 2^{x}. \]

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To find the exponential function \( y = a b^{x} \) that passes through the points \( (3, 56) \) and \( (5, 224) \), we can start by creating a system of equations using these points. Plugging in the point \( (3, 56) \): \[ 56 = a b^{3} \] (1) Plugging in the point \( (5, 224) \): \[ 224 = a b^{5} \] (2) Next, we can solve equations (1) and (2) simultaneously. Dividing equation (2) by equation (1) to eliminate \( a \): \[ \frac{224}{56} = \frac{a b^{5}}{a b^{3}} \] which simplifies to: \[ 4 = b^{2} \] So, we can take the square root of both sides: \[ b = 2 \quad \text{(since the base must be positive)} \] Now we can substitute \( b = 2 \) back into equation (1) to find \( a \): \[ 56 = a (2^{3}) \] \[ 56 = a \cdot 8 \] Now divide both sides by 8: \[ a = 7 \] So, we have \( a = 7 \) and \( b = 2 \). Therefore, the function is: \[ y = 7 \cdot 2^{x} \] Feel free to verify by plugging in the original points!

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