# Simplifying log. expressions.

• Jul 7th 2008, 07:12 PM
mankvill
Simplifying 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 PM
kalagota
Quote:

Originally Posted by mankvill
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

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 PM
mankvill
...I'm so confused.
• Jul 7th 2008, 07:58 PM
kalagota
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 PM
mankvill
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 PM
kalagota
aww.. first, be calm because stress will kill you.. (Wink)

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..

• Jul 7th 2008, 08:11 PM
mankvill
• Jul 7th 2008, 08:14 PM
kalagota
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 PM
Chop 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.

EDIT: Kalagota, didn't see that you posted this already. :o
• Jul 7th 2008, 08:19 PM
kalagota
Quote:

Originally Posted by Chop Suey
ln(a/bc) = ln(a) - ln(bc) = ln(a) - (ln(b) +ln(c)) = ln(a) - ln(b) - ln(c)

just added something in the middle on how to come on the last expression.
• Jul 7th 2008, 08:20 PM
kalagota
Quote:

Originally Posted by Chop Suey
EDIT: Kalagota, didn't see that you posted this already. :o

no problem. :)