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

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