A logarithmic function of the form \( y=a+b \ln (x-1) \) passes through the points \( (2,12) \) and \( (751,-30) \). Find \( a \) and \( b \), rounding values to three decimal places, if necessary. Answer Attempt 1 out of 2 \( a=\square b=\square \)

To find the values of \( a \) and \( b \) in the logarithmic function \( y = a + b \ln(x-1) \) that pass through the points \( (2,12) \) and \( (751,-30) \), we can substitute the given points into the function and solve the resulting system of equations. Given points: 1. \( (2,12) \) 2. \( (751,-30) \) Substitute the points into the function: 1. For point \( (2,12) \): \[ 12 = a + b \ln(2-1) \] \[ 12 = a + b \ln(1) \] \[ 12 = a + b \cdot 0 \] \[ 12 = a \] 2. For point \( (751,-30) \): \[ -30 = a + b \ln(751-1) \] \[ -30 = a + b \ln(750) \] Now, we have a system of equations: 1. \( 12 = a \) 2. \( -30 = a + b \ln(750) \) We can solve this system of equations to find the values of \( a \) and \( b \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}12=a\\-30=a+b\ln{\left(750\right)}\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}12=a\\-30=a+\ln{\left(750\right)}\times b\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}a=12\\-30=a+\ln{\left(750\right)}\times b\end{array}\right.\) - step3: Substitute the value of \(a:\) \(-30=12+\ln{\left(750\right)}\times b\) - step4: Swap the sides: \(12+\ln{\left(750\right)}\times b=-30\) - step5: Move the constant to the right side: \(\ln{\left(750\right)}\times b=-30-12\) - step6: Subtract the numbers: \(\ln{\left(750\right)}\times b=-42\) - step7: Divide both sides: \(\frac{\ln{\left(750\right)}\times b}{\ln{\left(750\right)}}=\frac{-42}{\ln{\left(750\right)}}\) - step8: Divide the numbers: \(b=-\frac{42}{\ln{\left(750\right)}}\) - step9: Calculate: \(\left\{ \begin{array}{l}a=12\\b=-\frac{42}{\ln{\left(750\right)}}\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}a=12\\b=-\frac{42}{\ln{\left(750\right)}}\end{array}\right.\) - step11: Rewrite: \(\left(a,b\right) = \left(12,-\frac{42}{\ln{\left(750\right)}}\right)\) The solution to the system of equations is: \[ a = 12 \] \[ b = -\frac{42}{\ln(750)} \] Rounding the values to three decimal places, we have: \[ a \approx 12 \] \[ b \approx -6.344 \] Therefore, the values of \( a \) and \( b \) are approximately \( a = 12 \) and \( b = -6.344 \).