Answer
The solutions are $x=-2$ and $x=3$.
Work Step by Step
RECALL:
$\log_a{b}=x \longrightarrow a^x=b$
where $b\gt0$
Use this rule in converting the equation to its equivalent exponential equation to have:
$6^1=x^2-x
\\6=x^2-x$
Subtract 6 to both sides to have:
$0=x^2-x-6
\\x^2-x-6=0$
Factor the trinomial to have:
$(x-3)(x+2)=0$
Equate each factor to zero then solve each equation to have:
$x-3=0 \text{ or } x+2 = 0
x=3 \text{ or } x=-2$
Note that in $\log_a{b}$, the value of $b$ has to be positive.
This means that in $\log_6{(x^2-x)}$, the value of $x^2-x$ must be positive.
Check if both solutions are valid to have:
For x = 3:
$x^2-x = 3^2 - 3 = 9-3=6$
For x=-2:
$x^2-x = (-2)^2 -(-2) = 4+2 = 6$
Since both solutions make $x^2-x$ positive, then both are valid.
Therefore, the solutions to the given equation are $-2$ and $3$.
