How do you solve #Ln(7-2x) = Ln(3x+3) # and find any extraneous solutions?

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Answer 1 · 2016-05-20T14:13:56

#ln(7 - 2x) - ln(3x + 3) = 0#

#ln((7 - 2x)/(3x + 3)) = 0#

#(7 - 2x)/(3x + 3) = e^0#

#(7 - 2x)/(3x + 3) = 1#

#7 - 2x = 1(3x + 3)#

#7 - 2x = 3x + 3#

#7 - 3 = 5x#

#4 = 5x#

#4/5 = x#

Hopefully this helps!