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

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

How to solve this equation?

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

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}

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

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

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

How to solve this equation?

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

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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}

Don't forget to check that $x>4$. - Galc127 yesterday

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