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# Confused Tweet

Maths and statistics discussion, revision, exam and homework help.

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1. Confused
So I am majorly confused atm..

If you have:

1) -ln|1-x| + ln|3+x|2, then you could use the rule to get:

-ln|(1-x)(3+x)2| right?

but if you re-arrange 1) to get ln|3+x|2 - ln|1-x|, then wouldn't you be able to get:

ln|(3+x)2/(1-x)|
2. Re: Confused
(Original post by sabre2th1)
So I am majorly confused atm..

If you have:

1) -ln|1-x| + ln|3+x|2, then you could use the rule to get:

-ln|(1-x)(3+x)2| right?

but if you re-arrange 1) to get ln|3+x|2 - ln|1-x|, then wouldn't you be able to get:

ln|(3+x)2/(1-x)|

Is there really a minus sign at the start?

If this is the equation, then you do the following,

Last edited by raheem94; 12-06-2012 at 00:00.
3. Re: Confused
(Original post by raheem94)
..
Yep, that is correct, there is a minus sign at the start..

So why can't I get:

-ln|(1-x)(3+x)2| ?

Thanks
4. Re: Confused
(Original post by sabre2th1)
So I am majorly confused atm..

If you have:

, then you could use the rule to get:

right?
Nope. The rule is . Or equivalently

Not . Remember, a sum has no particular order, therefore
Last edited by Lord of the Flies; 12-06-2012 at 00:34.
5. Re: Confused
(Original post by Lord of the Flies)
Nope. The rule is

Not . Remember, a sum has no particular order, therefore
Ah THANKS a lot ! Would the (a,b > 0) also apply when you are doing
ln a- ln b = ln a/b ?
6. Re: Confused
(Original post by sabre2th1)
Ah THANKS a lot ! Would the (a,b > 0) also apply when you are doing
ln a- ln b = ln a/b ?
Of course - for real numbers is only defined for . So whatever is in the logarithm must always be positive, hence the use of absolute values.

If either you have made a mistake, or you conclude the expression is undefined therefore [...], or you forgot to write it , or you are working outside of the real numbers...
Last edited by Lord of the Flies; 12-06-2012 at 00:42.
7. Re: Confused
(Original post by Lord of the Flies)
Of course - for real numbers is only defined for . So whatever is in the logarithm must always be positive, hence the use of absolute values.

If either you have made a mistake, or you conclude the expression is undefined therefore [...], or you forgot to write it , or you are working outside of the real numbers...
I understand! Thanks a lot