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

Feb 12, 2015

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

But now we do this the other way around:

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

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

We can factorise the top part:

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

Cancel out the $\left(x - 8\right)$'s

$\ln \left(x + 5\right)$
Domain : $x > 8$
(or else the original $\ln \left(x - 8\right)$ would be invalid)