How do you solve $ln (- 3 x - 1) - ln 7 = 2$ $ln(-3x-1)-ln7=2$ ?

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Answer 1 · 2016-07-25T07:34:52

$x = - \frac{7 e^{2} + 1}{3}$ $x=-(7e^2+1)/3$

Since $log a - log b = \frac{log a}{b}$ $loga-logb=loga/b$ ,

you can rewrite the equation as:

$ln (\frac{- 3 x - 1}{7}) = 2$ $ln((-3x-1)/7)=2$

you can note that $2 = {ln e}^{2}$ $2=lne^2$ , then

$ln (\frac{- 3 x - 1}{7}) = {ln e}^{2}$ $ln((-3x-1)/7)=lne^2$

or

$\frac{- 3 x - 1}{7} = e^{2} and - 3 x - 1 > 0$ $(-3x-1)/7=e^2 and -3x-1>0$

$- 3 x - 1 = 7 e^{2} and - 3 x > 1$ $-3x-1=7e^2 and -3x>1$

$- 3 x = 7 e^{2} + 1 and 3 x < - 1$ $-3x=7e^2+1 and 3x<-1$

$x = - \frac{7 e^{2} + 1}{3} and x < - \frac{1}{3}$ $x=-(7e^2+1)/3 and x<-1/3$

that's true

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