Confused
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ConfusedSo 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)| -
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Re: ConfusedIs this your first equation(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. -
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Re: ConfusedYep, that is correct, there is a minus sign at the start..(Original post by raheem94)
..
So why can't I get:
-ln|(1-x)(3+x)2| ?
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Re: ConfusedNope. The rule is(Original post by sabre2th1)
So I am majorly confused atm..
If you have:
, then you could use the rule to get:
right?
. Or equivalently 
Not
. Remember, a sum has no particular order, therefore 
Last edited by Lord of the Flies; 12-06-2012 at 00:34. -
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Re: ConfusedAh THANKS a lot(Original post by Lord of the Flies)
Nope. The rule is
Not. Remember, a sum has no particular order, therefore
! Would the (a,b > 0) also apply when you are doing
ln a- ln b = ln a/b ? -
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Re: ConfusedOf course - for real numbers(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 ?
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. -
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Re: ConfusedI understand! Thanks a lot(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...
! Would the (a,b > 0) also apply when you are doing 
