To solve the exercise you can use these rules of logarithms:
[tex]\begin{gathered} \log (\frac{m}{n})=\log (m)-\log (n)\Rightarrow\text{ Quotient rule} \\ \log (mn)=\log (m)+\log (n)\Rightarrow\text{Product rule} \end{gathered}[/tex]So, in this case, you have
[tex]\begin{gathered} \ln (\frac{2e}{x})=\ln (2e)-\ln (x) \\ \ln (\frac{2e}{x})=\ln (2)+\ln (e)-\ln (x) \\ \ln (\frac{2e}{x})=\ln (2)+1-\ln (x) \\ \text{ Because }\ln (e)=1 \\ \text{Now reorder} \\ \ln (\frac{2e}{x})=1+\ln (2)-\ln (x) \end{gathered}[/tex]Therefore, the expression that is equivalent to the given expression is
[tex]1+\ln (2)-\ln (x)[/tex]and the correct answer is option C.
