How do you solve #ln3+ln(2x^2+4)=ln12#?

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Answer 1 · 2016-08-10T16:57:25

x=0

Since #lna+lnb=ln(ab)#, then

#ln3+ln(2x^2+4)=ln12#

is equivalent to:

#ln(3(2x^2+4))=ln12#.

that's equivalent to:

#3(2x^2+4)=12#

#6x^2+12=12#

#6x^2=0#

#x=0#