Question

which expression is equivalent to the given expression? In(2e/x)A. In2+InxB.In1 + In2-InxC. 1+In2-InxD. In2-Inx

Answer 1

To solve the exercise you can use these rules of logarithms:

`\begin{gathered} \log ((m)/(n))=\log (m)-\log (n)\Rightarrow\text{ Quotient rule} \\ \log (mn)=\log (m)+\log (n)\Rightarrow\text{Product rule} \end{gathered}`

So, in this case, you have

`\begin{gathered} \ln ((2e)/(x))=\ln (2e)-\ln (x) \\ \ln ((2e)/(x))=\ln (2)+\ln (e)-\ln (x) \\ \ln ((2e)/(x))=\ln (2)+1-\ln (x) \\ \text{ Because }\ln (e)=1 \\ \text{Now reorder} \\ \ln ((2e)/(x))=1+\ln (2)-\ln (x) \end{gathered}`

Therefore, the expression that is equivalent to the given expression is

`1+\ln (2)-\ln (x)`

and the correct answer is option C.