How do you solve #ln(-4x-4)-ln3=3#?

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Answer 1 · 2016-08-18T03:55:54

Use the rule #lnx - lna = ln(x/a)#:

#=> ln((-4x - 4)/3) = 3#

#=> (-4x - 4)/3 = e^3#

#=> -4x - 4 = 3e^3#

#-4x = 3e^3 + 4#

#x = -1/4(3e^3 + 4)#

Hopefully this helps!