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

$x=-2$ and $x=4$
$\log_{8}(x^{2}-2x)=1$ Rewrite in exponential form: $8^{1}=x^{2}-2x$ $x^{2}-2x=8$ Take the $8$ to the left side of the equation": $x^{2}-2x-8=0$ Solve this equation by factoring: $(x+2)(x-4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+2=0$ $x=-2$ $x-4=0$ $x=4$ The initial equation is not undefined for either of the solutions found. The answer is $x=-2$ and $x=4$
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