ln(-3)
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ln(-3)Hi guys, if anybody could help me with the question in the title I'd be really grateful
I've ran through question 1 June 2011 C4 WJEC multiple times.
I keep ending up trying to calculate ln(-3)
Is it the same as -ln(3) or am I way off the mark
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Re: ln(-3)it is not the same sadly.(Original post by PinkyPurply)
Hi guys, if anybody could help me with the question in the title I'd be really grateful
I've ran through question 1 June 2011 C4 WJEC multiple times.
I keep ending up trying to calculate ln(-3)
Is it the same as -ln(3) or am I way off the mark
not seen the question but if you created the ln() function by integration remember that it is
ln(|arg|) that is created not just ln(arg) -
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Re: ln(-3)(Original post by PinkyPurply)
Hi guys, if anybody could help me with the question in the title I'd be really grateful
I've ran through question 1 June 2011 C4 WJEC multiple times.
I keep ending up trying to calculate ln(-3)
Is it the same as -ln(3) or am I way off the mark
May be you have made a mistake in the question can you give the link to the paper? -
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Re: ln(-3)You will have made an error earlier in your working; complex logarithms don't come up at A-level. As far as the A-level syllabus is concerned, logarithms of negative numbers are "undefined". What's the whole question?(Original post by PinkyPurply)
Hi guys, if anybody could help me with the question in the title I'd be really grateful
I've ran through question 1 June 2011 C4 WJEC multiple times.
I keep ending up trying to calculate ln(-3)
Is it the same as -ln(3) or am I way off the mark
Or(Original post by Lord of the Flies)
That only makes sense if you're working with complex numbers, in which case
.
or 
or 
or ...!
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Re: ln(-3)You have probably made a mistake somewhere.(Original post by PinkyPurply)
Hi guys, if anybody could help me with the question in the title I'd be really grateful
I've ran through question 1 June 2011 C4 WJEC multiple times.
I keep ending up trying to calculate ln(-3)
Is it the same as -ln(3) or am I way off the mark
For part (b) after integrating i get![\displaystyle \left[ \ln|x-3| + \frac3{x+2} \right]^7_6 = \left( \ln 4 + \frac13 \right) - \left( \ln 3 + \frac38 \right) = \ln \left( \frac43 \right) - \frac1{24}](http://www.thestudentroom.co.uk/latexrender/pictures/cf/cfd27aad977bc320af34f06e49e61207.png "\displaystyle \left[ \ln|x-3| + \frac3{x+2} \right]^7_6 = \left( \ln 4 + \frac13 \right) - \left( \ln 3 + \frac38 \right) = \ln \left( \frac43 \right) - \frac1{24}")
The first question of this paper.(Original post by nuodai)
What's the whole question?Last edited by raheem94; 12-06-2012 at 02:57. -
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Re: ln(-3)OP, you need to learn how to answer these questions for yourself.
The way you do this is by asking what the things actually mean.
What does lnx mean?
The definition is 'the number y such that e^y = x'.
So what does ln-3 mean? It means 'the number y such that e^y = -3'. But the [real] exponential function is never negative, so there can be no such number. So ln-3 does not exist.
-ln3 means 'find the number y such that e^y = 3 and then take the negative of it'. This time there is such a number y.
So obviously -ln3 cannot be the same as ln-3 because one exists and the other doesn't. -
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Re: ln(-3)(Original post by Xei)
I always wonder what kind of person it is who rates posts like my answer down.
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