Question

what is the domain of ln (2x^2 -4)-ln(x+1)-ln(x-1)

Answer 1

fine the places there the inner brackets equal 0
because ln (0) = DNE

Answer 2

domain of ln(x) is x>0 so you need the intersection of these three
2x^2-4>0 x+1>0 and x-1>0

Answer 3

the inner brackets of each expression must be > than 0

Answer 4

and then do what he says ^ lol

Answer 5

just make sure u give meee the MEDAL in the end! :D

Answer 6

i worked out the x values of each ln what is the next step the - between the ln confuses me

Answer 7

solve the 3 brackets where x>0 and see which of those x values will be valid for all of them

Answer 8

so if x = +/- 2 and +/- 1 how do we end the answer and what is the domain?

Answer 9

basically x must be greater than > sqrt2

Answer 10

x+1>0 and x-1>0 implies x>1 and x>-1 and 2x^2-4>0 imples x<-sqt(2) or x>sqrt(2)
so x>1 and x>-1 and (x>sqrt(2) or x<-sqr(2) so x>sqrt(2) is the domain

Answer 11

(-OO; 2) ? is the domain?

Answer 12

\[(\sqrt(2),infinity)\]

Answer 13

thank you

Answer 14

because x>1 and x>-1 so x cant be <-1

Answer 15

Well, \(\displaystyle\ln x= ln|x|+ i2\arctan\left(\frac{x-\overline x}{2i\sqrt{\left(\dfrac{x+\overline x}{2}\right)^2+\left(\dfrac{x-\overline x}{2i}\right)^2}+\dfrac{x+\overline x}{2}}\right)\) even when x<0.