Use regression to find the power function of the form \( y=a x^{p} \) that passes through the points \( (2,192) \) and \( (3,1458) \). \( a= \) \( p= \) Round your values to at least 4 decimal places. Question Help: D Post to forum Submit Question

To find the values of \( a \) and \( p \) for the power function \( y=a x^{p} \) that passes through the points \( (2,192) \) and \( (3,1458) \), we can set up a system of equations. First, we can express the two points in terms of the power function: 1. For the point \( (2,192) \): \[ 192 = a \cdot 2^{p} \] 2. For the point \( (3,1458) \): \[ 1458 = a \cdot 3^{p} \] Now, we can solve for \( a \) in both equations: From the first equation: \[ a = \frac{192}{2^{p}} \] From the second equation: \[ a = \frac{1458}{3^{p}} \] Since both expressions equal \( a \), we can set them equal to each other: \[ \frac{192}{2^{p}} = \frac{1458}{3^{p}} \] Cross-multiplying gives us: \[ 192 \cdot 3^{p} = 1458 \cdot 2^{p} \] Dividing both sides by 192: \[ 3^{p} = \frac{1458}{192} \cdot 2^{p} \] Calculating \( \frac{1458}{192} \): \[ \frac{1458}{192} = 7.59375 \] Thus, the equation becomes: \[ 3^{p} = 7.59375 \cdot 2^{p} \] To solve for \( p \), we can take the logarithm of both sides: \[ \log(3^{p}) = \log(7.59375 \cdot 2^{p}) \] Using the properties of logarithms, we split the right-hand side: \[ p \log(3) = \log(7.59375) + p \log(2) \] Rearranging gives us: \[ p \log(3) - p \log(2) = \log(7.59375) \] \[ p (\log(3) - \log(2)) = \log(7.59375) \] Finally, we can solve for \( p \): \[ p = \frac{\log(7.59375)}{\log(3) - \log(2)} \approx \frac{0.8808}{0.4771 - 0.3010} \approx \frac{0.8808}{0.1761} \approx 5.0002 \] Now we can substitute \( p \) back into one of the original equations to find \( a \): Using \( p \approx 5 \) in the first equation: \[ 192 = a \cdot 2^{5} \implies 192 = a \cdot 32 \] \[ a = \frac{192}{32} = 6 \] Thus, the values rounded to at least four decimal places are: \[ a \approx 6.0000, \quad p \approx 5.0000 \] So the final answer is \( a \approx 6.0000 \) and \( p \approx 5.0000 \).