How do you write $ln (x^{2} - 3 x - 40) - 1 n (x - 8)$ $ln(x^2-3x-40)-1n(x-8)$ as a single logarithm?

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Answer 1 · 2015-02-12T12:22:44

Remember than $ln (\frac{a}{b}) = ln a - ln b$ $ln (a/b)=ln a-ln b$

But now we do this the other way around:

$ln (x^{2} - 3 x - 40) - ln (x - 8) =$ $ln(x^2-3x-40)-ln(x-8)=$

$ln (\frac{x^{2} - 3 x - 40}{x - 8}) =$ $ln((x^2-3x-40)/(x-8))=$

We can factorise the top part:

$ln (\frac{(x - 8) (x + 5)}{(x - 8)}) =$ $ln(((x-8)(x+5))/((x-8)))=$

Cancel out the $(x - 8)$ $(x-8)$ 's

Answer:

$ln (x + 5)$ $ln(x+5)$

Extra :
Domain : $x > 8$ $x>8$
(or else the original $ln (x - 8)$ $ln(x-8)$ would be invalid)

How do you write ln(x^2-3x-40)-1n(x-8)ln(x2−3x−40)−1n(x−8)ln(x^2-3x-40)-1n(x-8) as a single logarithm?