# Write the Expression as a Logarithm of a Single Quantity

• Oct 7th 2007, 12:50 PM
liz155
Write the Expression as a Logarithm of a Single Quantity
$\displaystyle \frac{3}{2}$ $\displaystyle ln(x^2+1)-ln(x+1)-ln(x-1)$

I know that $\displaystyle ln\frac{a}{b}=lna-lnb$

But I don't know which part to divide first. Is there any way to know what part to divide first or is this somehow a trick question?

Thanks so much!!!
• Oct 7th 2007, 12:54 PM
Jhevon
Quote:

Originally Posted by liz155
$\displaystyle \frac{3}{2}$ $\displaystyle ln(x^2+1)-ln(x+1)-ln(x-1)$

I know that $\displaystyle ln\frac{a}{b}=lna-lnb$

But I don't know which part to divide first. Is there any way to know what part to divide first or is this somehow a trick question?

Thanks so much!!!

it doesn't matter. you will get the same thing in the end (just mind the signs in front of the logs). also note that $\displaystyle \ln a + \ln b = \ln ab$ and $\displaystyle n \ln a = \ln \left(a^n \right)$
• Oct 7th 2007, 01:02 PM
liz155
Sorry I meant to write this...

$\displaystyle \frac{3}{2}$ [$\displaystyle ln(x^2+1)-ln(x+1)-ln(x-1)$]

Do the parenthesis make a difference?
So when I combine the stuff inside the parenthesis, I put the 3/2 as an exponent?

ln[($\displaystyle x^2$+1)(x-1)/(x+1)]^(3/2)

Does that seem right?

• Oct 7th 2007, 01:14 PM
Jhevon
Quote:

Originally Posted by liz155
Sorry I meant to write this...

$\displaystyle \frac{3}{2}$ [$\displaystyle ln(x^2+1)-ln(x+1)-ln(x-1)$]

Do the parenthesis make a difference?
So when I combine the stuff inside the parenthesis, I put the 3/2 as an exponent?

ln[($\displaystyle x^2$+1)(x-1)/(x+1)]^(3/2)

Does that seem right?

yes, the parentheses make a big difference. and you are incorrect. look at the rules and try gain. i told you to mind the signs in front of the logs, the middle log has a minus sign in front (you implicitly treated it as a plus sign)
• Oct 7th 2007, 01:24 PM
liz155
I see what I did wrong. Thank you!:)

I got...

ln(($\displaystyle x^2$+1)/($\displaystyle x^2$-1))^(3/2)

Does this seem right?

Thanks!:)
• Oct 7th 2007, 01:53 PM
Jhevon
Quote:

Originally Posted by liz155
I see what I did wrong. Thank you!:)

I got...

ln(($\displaystyle x^2$+1)/($\displaystyle x^2$-1))^(3/2)

Does this seem right?

Thanks!:)

correct