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?
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but plugging your solution in the equation we get $$-4\frac{1}{6}-4<0$$ – Dr. Sonnhard Graubner Dec 17 '16 at 20:53
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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).
Alex M.
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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.
Arnaldo
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