Question

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

Answer 1

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

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

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

${\log }_{10}\left(\frac{(x-3)(x+2)}{x+2}\right)=x-4$

${\log }_{10}(x-3)=x-4$ if $x\ne -2$

$\Rightarrow {10}^{(x-4)}=x-3$

Let $x-3=y$

then

${10}^{y-1}=y$

$\frac{{10}^y}{10}=y$

${10}^y=10y$

$\Rightarrow y=1$

$x-3=1$

$x=4$