# Math Help - Write the Expression as a Logarithm of a Single Quantity

1. ## Write the Expression as a Logarithm of a Single Quantity

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

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

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!!!

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

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

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 $\ln a + \ln b = \ln ab$ and $n \ln a = \ln \left(a^n \right)$

3. Sorry I meant to write this...

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

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

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

Does that seem right?

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

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

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

ln[( $x^2$+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)

5. I see what I did wrong. Thank you!

I got...

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

Does this seem right?

Thanks!

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

I got...

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

Does this seem right?

Thanks!
correct