Solution assignment 10 Logarithmic equations

Solve:

$\ln(x+2)+\ln(x-3)=\ln(x-1)+\ln(x+4)$

The equation is only defined when all logarithms are defined. This is the case when the following conditions are met:

$x>-2$ and $x>3$ and $x>1$ and $x>-4$

thus if:

$x>3$

We can write the equation as:

$\ln(x^2-x-6)=\ln(x^2+3x-4)$

$x^2-x-6=x^2+3x-4$

$-x-6=3x-4$

$x=-\displaystyle\frac{1}{2}$

This solution falls outside the interval given above and thus the equation has no solution.