How do you solve $ln (- 4 x - 4) - ln 3 = 3$ $ln(-4x-4)-ln3=3$ ?

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Answer 1 · 2016-08-18T03:55:54

Use the rule $ln x - ln a = ln (\frac{x}{a})$ $lnx - lna = ln(x/a)$ :

$\Rightarrow ln (\frac{- 4 x - 4}{3}) = 3$ $=> ln((-4x - 4)/3) = 3$

$\Rightarrow \frac{- 4 x - 4}{3} = e^{3}$ $=> (-4x - 4)/3 = e^3$

$\Rightarrow - 4 x - 4 = 3 e^{3}$ $=> -4x - 4 = 3e^3$

$- 4 x = 3 e^{3} + 4$ $-4x = 3e^3 + 4$

$x = - \frac{1}{4} (3 e^{3} + 4)$ $x = -1/4(3e^3 + 4)$

Hopefully this helps!

How do you solve ln(−4x−4)−ln3=3ln(-4x-4)-ln3=3?