Natural logarithms

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phy

i need help with this one question; we have to write ln9-3ln(squareroot 3) + ln81 in the form kln3 where k is an exact fraction; i dont remember how to do this at all so any help would be greatly appreciated; thanks
 

arildno

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Remember the general rule : ln(a^(b))=b*ln(a)
 

phy

so i could rewrite it as 2ln3-3ln(sqrt3)+4ln3; but then what?
 
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Remember that taking the square root is the same as raising to the 1/2 power. You may also want to combine the logs before making the exponents into the coefficients. Do you remember your log properties?
 

phy

um, not really
 

arildno

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Use Moose352 suggestion of sqrt(3)=3^(1/2). You will now have ln(3) as a common factor.
 

phy

so would the answer just be 4(4/3)ln3?
 

arildno

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How did you get the 1/3 denominator????
 

phy

well i said that 3ln(sqrt3) is the same as 3ln(3^1/2) which is the same as 3/2(ln3)
 

arildno

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I'm with you there, that's correct.
But it's 2 in the denominator not 3!
 

phy

yeah so dont we do 3*1/2 which is 3/2? i'll show you what i've done so far.
2ln3-3/2ln3+4ln3
(4ln3)(2ln3)/(3/2ln3)
4(4/3)ln3
 

arildno

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Allright, I see were you have mixed up:

Sum rule logarithm: ln(x)+ln(y)=ln(x*y)
This is not what you have done.
Exponent rule logarithms: ln(a^(b))=b*ln(a)

If you want to do it with the sum rule, write:

2ln(3)-3/2ln(3)+4ln(3)=ln(3^(2))+ln(1/(3^(3/2)))+ln(3^(4))=
ln(3^(2-3/2+4))=ln(3^(9/2))=9/2ln(3)

I'll check up on this thread tomorrow..
 

phy

ok i get it now; i see where i made my mistake. thanks a lot for your help :)
 

arildno

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A good advice:
When doing maths you are not too familiar with, keep the definitions right in front of you while you're working, until you don't need to look at them anymore.
 

phy

yeah, that is good advice; it's just that that was a review question from my calculus textbook, stuff that we learned in highschool and haven't seen in some time; we were just expected to remember how to do it and i forgot; well, thanks for all your help
 

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