Answer:
[tex]\ln{(3x)}+\ln{(x+1)}-2\ln{(2x+1)}[/tex]
Step-by-step explanation:
we need to keep in mind two properties of log:
[tex]\ln{\dfrac{3x(x+1)}{(2x+1)^2}}[/tex]
[tex]\ln{\left(\dfrac{3x(x+1)}{(2x+1)^2}\right)}\\\ln{(3x(x+1))}-\ln{((2x+1)^2)}\\\ln{(3x)}+\ln{(x+1)}-\ln{((2x+1)^2)}[/tex]
[tex]\ln{(3x)}+\ln{(x+1)}-2\ln{(2x+1)}[/tex]
this is the rewritten expression!