We will revise some rules of ln
[tex]\begin{gathered} m\ln (b)=\ln (b)^m\rightarrow(1) \\ \ln a+\ln b-\ln c=\ln \frac{ab}{c}\rightarrow(2) \end{gathered}[/tex]
In the given expression
[tex]3\ln a-\frac{1}{2}(\ln b+\ln c^2)[/tex]
By using the rule (1)
[tex]3\ln a-\frac{1}{2}(\ln b+\ln c^2)=\ln a^3-(\ln b+\ln c^2)^{\frac{1}{2}}[/tex]
By using rule (2) in the second term
[tex]\ln a^3-(\ln b+\ln c^2)^{\frac{1}{2}}=\ln a^3-\ln (bc^2)^{\frac{1}{2}}[/tex]
By using the rule of the exponent on b and c
[tex](bc^2)^{\frac{1}{2}}=(b^{\frac{1}{2}})(c^{2\times\frac{1}{2}})=b^{\frac{1}{2}}c=c\sqrt[]{b}[/tex]
The expression is
[tex]\ln a^3-\ln (bc^2)^{\frac{1}{2}}=\ln a^3-\ln c\sqrt[]{b}[/tex]
By using rule (2), then
[tex]\ln a^3-\ln c\sqrt[]{b}=\ln (\frac{a^3}{c\sqrt[]{b}})[/tex]
The answer is D
