Simplifying log. expressions.

Printable View

July 7th 2008, 07:12 PM
mankvill

Simplifying log. expressions.

Help. :( In a hurry.

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

I dunno how to do the math tags exactly, but the fraction is 15x^3 over y^5
July 7th 2008, 07:27 PM
kalagota

Quote:

Originally Posted by mankvill

Help. :( In a hurry.

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

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

did you mean $\ln \left(\frac{15x^3}{y^5}\right)$ ?

well, you can use this..

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

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

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

this must be sufficient.. pls post what you have done.. Ü
July 7th 2008, 07:29 PM
mankvill

...I'm so confused.
July 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..

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

can you do the rest? just give us what you already did..
July 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
July 7th 2008, 08:09 PM
kalagota

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 $\ln (5x^3) - \ln (y^5)$

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

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

can you follow it?
July 7th 2008, 08:11 PM
mankvill

i can't follow anything. (Headbang)
July 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?
July 7th 2008, 08:15 PM
Chop Suey

Mankvill: let's have a crash course in logarithms.

$\ln{ab} = \ln{a} + \ln{b}$
If you got multiplication within the ln brackets, you can separate them like that.

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

Another thing, you can also separate the ln if it had division within brackets:
$\ln{\frac{a}{b}} = \ln{a} - \ln{b}$

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

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
July 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.
July 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. :)