How to solve this equation? Can I treat as a quadratic equation?

Question

$$\ln(x+3)+\ln(x-4)=0$$

How to solve this equation?

First removing the 'ln' from the equation and after making a quadratic equation and then solve the quadratic equation?

Answer 1

Yes, more or less, if you meant the following:

$$\begin{align} \ln(x+3) + \ln(x - 4) = 0 & \iff \ln((x+3)(x-4)) = 0 \\ \\ &\iff (x+3)(x-4) = 1,\text{ and }x>4\\ \\ & \iff x^2 - x - 13 = 0, \text{ and }x\gt 4 \\ \\ &\iff\cdots\end{align}$$

Answer 2

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