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

1 Answer
Jul 25, 2016

Answer:

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

Explanation:

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