Question

${\log }_{10}(x^2-x-6)-x={\log }_{10}(x+2)-4$

Answer 1

The equation can be written as
$lo{g}_{10}\left(\frac{x^2-x-6}{x+2}\right)=x-4$
or $lo{g}_{10}\left(\frac{(x-3)(x+2)}{x+2}\right)=x-4$
or $lo{g}_{10}(x-3)=x-4$
or $x-3={10}^{x-4}$
By trial, x = 4 is the solution of (1) which is, therefore, the solution of the given equation.
Observe that x > 4 or x < 4 does not satisfy (1).
Hence x = 4 is the only solution of the original equation.