TITLE: How to solve this equation? Can I treat as a quadratic equation?
QUESTION [5 upvotes]: $$\ln(x+3)+\ln(x-4)=0$$
How to solve this equation?
First removing the 'ln' from the equation and after making a quadratic equation and then solve the quadratic equation?

REPLY [12 votes]: Yes, more or less, if you meant the following: 
$$\begin{align} \ln(x+3) + \ln(x - 4) = 0 & \iff \ln((x+3)(x-4)) = 0 \\ \\ &\iff (x+3)(x-4) = 1,\text{ and }x>4\\ \\ & \iff x^2 - x - 13 = 0, \text{ and }x\gt 4 \\ \\ &\iff\cdots\end{align}$$

REPLY [3 votes]: using the law of logarithm we have
$\ln((x+3)(x-4))=\ln(1)$
thus we have
$(x+3)(x-4)=1$