Find the equation of a power function of the form \( y=a x^{p} \) that passes through the points \( (4,60.8) \) and \( (3,34.2) \). Indicate your values for \( a \) and \( p \) below. Round to 4 decimal places as needed. \( a=\square \) \( p= \)

To find the equation of the power function \( y = ax^p \) that passes through the points \( (4, 60.8) \) and \( (3, 34.2) \), we will start by setting up two equations based on these points. 1. From the point \( (4, 60.8) \): \[ 60.8 = a \cdot 4^p \quad (1) \] 2. From the point \( (3, 34.2) \): \[ 34.2 = a \cdot 3^p \quad (2) \] Now we can solve these two equations simultaneously. From equation (1), we can express \( a \) in terms of \( p \): \[ a = \frac{60.8}{4^p} \] Substituting this expression for \( a \) into equation (2): \[ 34.2 = \left( \frac{60.8}{4^p} \right) \cdot 3^p \] This simplifies to: \[ 34.2 = 60.8 \cdot \frac{3^p}{4^p} \] Rearranging gives: \[ \frac{34.2}{60.8} = \left( \frac{3}{4} \right)^p \] Calculating \( \frac{34.2}{60.8} \): \[ \frac{34.2}{60.8} \approx 0.5625 \] Thus we have: \[ 0.5625 = \left( \frac{3}{4} \right)^p \] Taking the logarithm on both sides: \[ \log(0.5625) = p \cdot \log\left(\frac{3}{4}\right) \] Now we can solve for \( p \): \[ p = \frac{\log(0.5625)}{\log\left(\frac{3}{4}\right)} \] Calculating \( p \): \[ p \approx \frac{-0.2491}{-0.1249} \approx 1.994 \] Now substituting \( p \) back into equation (1) to find \( a \): \[ a = \frac{60.8}{4^{1.994}} \] \[ 4^{1.994} \approx 15.870 \] \[ a = \frac{60.8}{15.870} \approx 3.8281 \] Thus the final values are: \[ a \approx 3.8281 \] \[ p \approx 1.9940 \] So, rounding them to four decimal places gives: \( a= 3.8281 \) \( p= 1.9940 \)