How do you solve #ln(x^2+12)=lnx+ln8#?

1 Answer
Narad T.
Oct 17, 2016

#x=2 # and #x=6#

Explanation:

#ln(x^2+12)=lnx+ln8#
We can rewrite this as #ln(x^2+12)=ln8x#
Removing the ln on both sides,give
#x^2+12=8x#
Rearranging
#x^2-8x+12=0#
#(x-6)(x-2)=0#
so x=6 or x=2
We have to check the conditions for ln
#x^2+12>0#and #8x>0#
So the solution is ok

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