Inactive Tutor answered 07/17/16
Tutor
New to Wyzant
Tamika,
Is the problem to expand these in terms of simpler logarithms?
Assuming yes, and also assuming the groupings as shown, here is how I would do them.
ln (3x/(x+1)) = ln (3x) − ln (x+1) = ln 3 + ln x − ln (x+1).
log (√x (x+1)/x2) = log √x + log (x+1) − log (x2) = ½ log x + log (x+1) − 2 log x = -3/2 log x + log (x+1).
(I combined two log x terms together in the last step.)
Let me know if this is helpful or not. In the future, please put sufficient parentheses around expressions such as exponents and denominators to avoid confusion. It will get you the help quicker.