## Algebra: A Combined Approach (4th Edition)

$x=3$ and $x=-2$
$\log_{6}(x^{2}-x)=1$ Rewrite in exponential form: $6^{1}=x^{2}-x$ $x^{2}-x=6$ Take the $6$ to the left side of the equation: $x^{2}-x-6=0$ Solve this equation by factoring: $(x-3)(x+2)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-3=0$ $x=3$ $x+2=0$ $x=-2$ The initial equation is not undefined for either of the solutions, so our answer is: $x=3$ and $x=-2$