# Thread: Simplify natural log expression

1. ## Simplify natural log expression

Hi,

I'm asked to simplify this expression:

$\displaystyle ln(\frac{x^2 - y^2}{\sqrt{x}})$

Using log rules, I did this:

$\displaystyle ln(x^2 - y^2) - ln(\sqrt{x})$

$\displaystyle ln(x^2 - y^2) - \frac{1}{2}ln(x)$

Is it possible to simplify it further?

2. Originally Posted by centenial
Hi,

I'm asked to simplify this expression:

$\displaystyle ln(\frac{x^2 - y^2}{\sqrt{x}})$

Using log rules, I did this:

$\displaystyle ln(x^2 - y^2) - ln(\sqrt{x})$

$\displaystyle ln(x^2 - y^2) - \frac{1}{2}ln(x)$

Is it possible to simplify it further?
No. There no simplification for log(a+ b) or log(a- b) so that cannot be simplified further.

3. Thanks!

4. Hello, centenial!

I'm asked to simplify this expression:

$\displaystyle \ln\left(\frac{x^2 - y^2}{\sqrt{x}}\right)$

Using log rules, I did this:

. . $\displaystyle \ln(x^2 - y^2) - \ln(\sqrt{x})$

. . $\displaystyle \ln(x^2 - y^2) - \tfrac{1}{2}\ln(x)$

Is it possible to simplify it further?

HallsofIvy is correct!

But if you want to show off, we can factor and separate the first log:

. . $\displaystyle \ln\bigg[(x-y)(x+y)\bigg] - \tfrac{1}{2}\ln(x) \;\;=\;\;\ln(x-y) + \ln(x+y) - \tfrac{1}{2}\ln(x)$

I suspect that this is the answer "they" expect.
I asked myself, "Why did they use a factorable numerator?"