Solving for x $\ln{(-4x-2)}-\ln{(-4x)} = \ln{(4)}$

Question

I am trying to solve for x: $$\ln{(-4x-2)}-\ln{(-4x)} = \ln{(4)}$$ My attempt $$\ln{\left(\frac{-4x-2}{-4x}\right)} = \ln{(4)}$$ $$\frac{4x+2}{4x} = 4$$ $$4x+2 = 16x$$ $$2 = 12x$$ $$x = \frac{1}{6}.$$ Why does my solution not work? Is there even a solution?

Answer 1

Your calculation is right but you can't forget the boundary for the problem:

$$-4x-2>0 \Rightarrow x<-\frac{1}{2}$$

and

$$-4x>0 \Rightarrow x<0$$

Once your solution doesn't respect that boundary it means that the problem has no solution.

Answer 2

Your solution is not good because in order for $\ln (-4x)$ to exist one must have $x<0$; a similar condition imposed for $\ln (-4x-2)$ leads to $x < -\frac 1 2$, so intersecting these two conditions gives you $x < -\frac 1 2$. Since $\frac 1 6 > 0$, this equation has no solution (as you seem to have suspected).