Solve for \( y \) in terms of \( x \) \( \ln (y-5)-\ln 9=x+\ln x \) \( y=\square \)

To solve for \( y \) in terms of \( x \), we start with the equation: \[ \ln (y-5) - \ln 9 = x + \ln x \] Using the logarithmic property that states \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \), we can rewrite the left side: \[ \ln \left(\frac{y-5}{9}\right) = x + \ln x \] Next, we can exponentiate both sides to eliminate the natural logarithm: \[ \frac{y-5}{9} = e^{x + \ln x} \] Recognizing that \( e^{x + \ln x} = e^x \cdot x \), we rewrite it as: \[ \frac{y-5}{9} = x e^x \] Now, we can solve for \( y \) by multiplying both sides by 9 and then adding 5: \[ y - 5 = 9 x e^x \] Finally, adding 5 gives us: \[ y = 9 x e^x + 5 \] So the final expression for \( y \) in terms of \( x \) is: \[ y = 9 x e^x + 5 \]