# How do you expand ln((6x) / sqrt(x^2 – 4))?

$\ln \left(6\right) + \ln \left(x\right) - \left(\frac{1}{2}\right) \left(\ln \left(x + 2\right) + \ln \left(x - 2\right)\right)$

#### Explanation:

$\ln \left(\frac{6 x}{\sqrt{{x}^{2} - 4}}\right)$

The first we can do to expand this is to have ln of the numerator - ln of the denominator:

$\ln \left(6 x\right) - \ln \sqrt{{x}^{2} - 4}$

I'm going to rewrite the square root into an exponent:

$\ln \left(6 x\right) - \ln {\left({x}^{2} - 4\right)}^{\frac{1}{2}}$

I can now move the 1/2 to in front of the ln:

$\ln \left(6 x\right) - \left(\frac{1}{2}\right) \ln \left({x}^{2} - 4\right)$

I can also factor the ${x}^{2} - 4$ term:

$\ln \left(6 x\right) - \left(\frac{1}{2}\right) \ln \left(\left(x + 2\right) \left(x - 2\right)\right)$

which lends itself to:

$\ln \left(6 x\right) - \left(\frac{1}{2}\right) \left(\ln \left(x + 2\right) + \ln \left(x - 2\right)\right)$

and I almost forgot that I can do the same to the 6x:

$\ln \left(6\right) + \ln \left(x\right) - \left(\frac{1}{2}\right) \left(\ln \left(x + 2\right) + \ln \left(x - 2\right)\right)$