Question 12 - of 21 step 1 of 1 Use the properties of logarithms to condense the following expression as much as possible, writing the answer as a single term with a coefficient of. 1 . All exponents should be positive. \[ 5\left(\ln \left(\sqrt[3]{e^{7}}\right)-\ln (x y)\right) \]

First, let’s break it down! We know that \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\). So, we can use this property to combine the logarithms: \[ \ln \left( \frac{\sqrt[3]{e^{7}}}{xy} \right). \] Next, we simplify \(\sqrt[3]{e^{7}}\) as \(e^{\frac{7}{3}}\). Thus, our expression becomes: \[ \ln \left( \frac{e^{\frac{7}{3}}}{xy} \right). \] Using the logarithm property \(\ln(a^b) = b \cdot \ln(a)\), we can condense even further while considering the coefficient of 5 that is outside: \[ 5 \cdot \ln \left( \frac{e^{\frac{7}{3}}}{xy} \right) = 5 \left( \ln (e^{\frac{7}{3}}) - \ln(xy) \right). \] This leads us to \(5 \cdot \frac{7}{3} = \frac{35}{3}\), so it becomes: \[ \ln \left( \frac{e^{\frac{7}{3}}}{xy} \right)^5 = \ln\left( \frac{e^{\frac{35}{3}}}{x^5y^5} \right). \] Finally, the condensed expression is: \[ \ln \left( \frac{e^{\frac{35}{3}}}{x^5y^5} \right). \]