Question

how come the domain of ln(x)-ln(x-1)=y isn't the same with ln(x/(x-1)) ?

http://www.wolframalpha.com/input/?i=ln%28x%2F%28x-1%29%29
and
http://www.wolframalpha.com/input/?i=ln%28x%29+-+ln%28x-1%29&dataset

Answer 1

@.Sam. @hartnn @estudier

Answer 2

knowing that you can combine the ln's in ln(x)-ln(x-1) to ln(x/(x-1)) right?

Answer 3

I'd say it is the same tbh.

Answer 4

Trying to open the wolf pages now, it's being very slow.

Answer 5

you can type the expressions and it'll give out the domain in case it's taking really long:))

Answer 6

What does it say the domain for each is?

Answer 7

I can't even open wolf for some reason :S

Answer 8

for 1st, x>1
for 2nd, x<0 or x>1
i would go with 1st only.

Answer 9

{x element R : x>1} for ln(x)-ln(x-1)
{x element R : x<0 or x>1} for ln(x/(x-1))

Answer 10

is it just because in case there's a negative number in the numerator and denominator of x/(x-1), it can cancel out and the expression would be positive??

Answer 11

Ah I see, both equations are equivalent, you can write the first as the second.

{x element R : x<0 or x>1} is the more complete answer when you write the equation in the combined way, so I would go with that.

Answer 12

buuuuuuuut.... x can't be a negative number for the first expression?

Answer 13

True, but when the equation is put in its simplest form \(x\) can take a negative number.

Answer 14

what do you guys think? @.Sam. ? :D


@hartnn ? :D


@estudier ? :D

etc?

Answer 15

which equation are ya talking about traxter? :D

Answer 16

ln(x)-ln(x-1)=y can be written as y=ln(x/(x-1)).
The first equation, left in that form, can not take any negative values of \(x\).
However, if we have the function \(y=f(x)=ln(x)-ln(x-1)\) then we can simplify this to \(y=f(x)=ln(\frac{x}{x-1})\), which would be able to take x<0 or x>1.

Answer 17

so do you think that when getting a domain of an expression with logarithms, one shouldn't combine the logs first?

Answer 18

Put x = -1

Answer 19

what do you mean, @estudier ?

Answer 20

Well ln 1/2 is OK, but not ln -1

Answer 21

U set the domain...

Answer 22

domain of y = x/x is not same as y = 1

Answer 23

You don't find a domain, you set the domain, it is part of the definition of a function..

Answer 24

@ganeshie8 makes a good point.
The domain just tells you which values of \(x\) you aren't allowed to put into the function.
If you left the equation as it was without combining the logs, you would have to have domain as \(x>1\).

Answer 25

thanks!!! now i really get it! someone please give a medal to @ganeshie8 too!