## Intermediate Algebra (6th Edition)

$x=-1$ and $x=4$
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$. Therefore, $log_{4}(x^{2}-3x)=1$ means $4^{1}=4=x^{2}-3x$. Subtract 4 from both sides of the equation to get all terms on one side. $x^{2}-3x-4=0$ We know that 1 and 4 are factors of 4 (and the sum of 1 and -4 is -3, which is the coefficient attached to the middle term). Therefore, we can factor this equation into $(x+1)(x-4)=0$. After setting both terms equal to 0, we know that $x=-1$ and $x=4$.