How do you solve #ln(-3x-1)-ln7=2#?

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Answer 1 · 2016-07-25T07:34:52

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

Since #loga-logb=loga/b#,

you can rewrite the equation as:

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

you can note that #2=lne^2#, then

#ln((-3x-1)/7)=lne^2#

or

#(-3x-1)/7=e^2 and -3x-1>0#

#-3x-1=7e^2 and -3x>1#

#-3x=7e^2+1 and 3x<-1#

#x=-(7e^2+1)/3 and x<-1/3#

that's true