What is the domain and range of #y = ln((2x-1)/(x+1))#?

1 Answer
Jan 20, 2016


Domain is set of all positive real numbers greater than #1/2#
Range is the entire real number system.


Given log functions can take values that are either above 0 or below infinite, basically the positive side of the Real number axis.
So, #log(x)inRR " "AA x in RR^+#
Here, #x " is simply " (2x-1)/(x+1)#
So, #(2x-1)/(x+1)>0\impliesx!=0" "x>1/2#
Of course, the range of the log function is the entire real number system.

Note in the above answer, I did not regard the complex numbers at all.