How do you write #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 (a/b)=ln a-ln b#

But now we do this the other way around:

#ln(x^2-3x-40)-ln(x-8)=#

#ln((x^2-3x-40)/(x-8))=#

We can factorise the top part:

#ln(((x-8)(x+5))/((x-8)))=#

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

Answer:

#ln(x+5)#

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