natural logarithms

[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]

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

[phy]

so i could rewrite it as 2ln3-3ln(sqrt3)+4ln3; but then what?

[Moose352]

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]

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]

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]

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]

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]

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