Simplify natural log expression

**centenial** · February 8th 2010, 06:28 AM

Hi,

I'm asked to simplify this expression:

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

Using log rules, I did this:

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

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

Is it possible to simplify it further?

**HallsofIvy** · February 8th 2010, 06:42 AM

Originally Posted by centenial

Hi,

I'm asked to simplify this expression:

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

Using log rules, I did this:

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

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

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.

**centenial** · February 8th 2010, 06:46 AM

Thanks!

**Soroban** · February 8th 2010, 07:17 AM

Hello, centenial!

I'm asked to simplify this expression:

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

Using log rules, I did this:

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

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

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:

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

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

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