Help. :( In a hurry.

$\displaystyle ln(15x^3)/y^5$

I dunno how to do the math tags exactly, but the fraction is 15x^3 over y^5

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- Jul 7th 2008, 07:12 PMmankvillSimplifying log. expressions.
Help. :( In a hurry.

$\displaystyle ln(15x^3)/y^5$

I dunno how to do the math tags exactly, but the fraction is 15x^3 over y^5 - Jul 7th 2008, 07:27 PMkalagota
did you mean $\displaystyle \ln \left(\frac{15x^3}{y^5}\right)$?

well, you can use this..

$\displaystyle \ln (ab) = \ln a + \ln b$

$\displaystyle \ln (a/b) = \ln a - \ln b$

$\displaystyle \ln (a^b) = b\ln a$

this must be sufficient.. pls post what you have done.. Ü - Jul 7th 2008, 07:29 PMmankvill
...I'm so confused.

- Jul 7th 2008, 07:58 PMkalagota
aww..

i'll do the first step..

notice that inside the parenthesis is a ratio.. so you can use the second identity i've given you..

$\displaystyle \ln \left(\frac{5x^3}{y^5}\right) = \ln (5x^3) - \ln (y^5)$

can you do the rest? just give us what you already did.. - Jul 7th 2008, 07:59 PMmankvill
I honestly have no idea, sorry. I'm drawing a blank. I'm under so much stress right now. x_x

- Jul 7th 2008, 08:09 PMkalagota
aww.. first, be calm because stress will kill you.. (Wink)

next, read your notes and recall everything you already know..

i'll do this example completely.. try to study and try to answer the rest..

so, we are left with $\displaystyle \ln (5x^3) - \ln (y^5)$

$\displaystyle \underbrace{\ln 5x^3} - \ln y^5 = \underbrace{\ln 5 + \ln (x^3)} - \ln (y^5)$ i used the first identity i gave.. for the minuend.

then $\displaystyle \ln 5 + \ln (x^3) - \ln (y^5) = \ln 5 + 3 \ln x - 5 \ln y$ this time, i used the third identity..

can you follow it? - Jul 7th 2008, 08:11 PMmankvill
i can't follow anything. (Headbang)

- Jul 7th 2008, 08:14 PMkalagota
i already give the complete solution until to the final answer.. try reading it along with other examples from your books or sources..

i think, it's late night there. have rested already? - Jul 7th 2008, 08:15 PMChop Suey
Mankvill: let's have a crash course in logarithms.

$\displaystyle \ln{ab} = \ln{a} + \ln{b}$

If you got multiplication within the ln brackets, you can separate them like that.

Another example: $\displaystyle \ln{abc} = \ln{a} + \ln{b} + \ln{c}$

Another thing, you can also separate the ln if it had division within brackets:

$\displaystyle \ln{\frac{a}{b}} = \ln{a} - \ln{b}$

$\displaystyle \ln{\frac{a}{bc}} = \ln{a} - \ln{b} - \ln{c}$

Notice that when they're multiplied, they ln of each value inside is added. When they're divided, it becomes: ln of the numerator minus ln of the denominator.

Do you follow now? If you do, reread kalagota's answers.

EDIT: Kalagota, didn't see that you posted this already. :o - Jul 7th 2008, 08:19 PMkalagota
- Jul 7th 2008, 08:20 PMkalagota