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

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Answer 1 · 2016-05-28T14:23:10

$ln(6)+ln(x)-(1/2)(ln(x+2)+ln(x-2))$

Let's start with the original:

$ln((6x)/sqrt(x^2-4))$

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

$ln(6x)-ln sqrt(x^2-4)$

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

$ln(6x)-ln(x^2-4)^(1/2)$

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

$ln(6x)-(1/2)ln(x^2-4)$

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

$ln(6x)-(1/2)ln((x+2)(x-2))$

which lends itself to:

$ln(6x)-(1/2)(ln(x+2)+ln(x-2))$

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

$ln(6)+ln(x)-(1/2)(ln(x+2)+ln(x-2))$

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