Can someone explain why the domain for y=ln(x^2-1) is |x|>1
But why the domain for y = ln(x+1) + ln(x-1) is just x>1
Whut is going on.
2
comments:
Anonymous
said...
for the domain to work you need the thing inside the ln to be positive, so if you solve it for x^2-1 > 0 you get x<-1 or x>1, whereas if you solve both x+1 and x-1 > 0 you get x is necessarily greater than 1. the two appear to be the same function but the second function is just a subset of the first since the solution to y=ln(x^2-1) can be y=ln(x+1)+ln(x-1) (the branch x>1) and y=ln(1-x)+ln(-1-x) (the branch x<-1).
2 comments:
for the domain to work you need the thing inside the ln to be positive, so if you solve it for x^2-1 > 0 you get x<-1 or x>1, whereas if you solve both x+1 and x-1 > 0 you get x is necessarily greater than 1. the two appear to be the same function but the second function is just a subset of the first since the solution to y=ln(x^2-1) can be y=ln(x+1)+ln(x-1) (the branch x>1) and y=ln(1-x)+ln(-1-x) (the branch x<-1).
Kudos (Y)
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