Apply the formula: $a\log_{b}\left(x\right)$$=\log_{b}\left(x^a\right)$

The difference of two logarithms of equal base $b$ is equal to the logarithm of the quotient: $\log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\right)$

For two logarithms of the same base to be equal, their arguments must be equal. In other words, if $\log(a)=\log(b)$ then $a$ must equal $b$

Move everything to the left hand side of the equation

Factor the trinomial $x^2-x-6$ finding two numbers that multiply to form $-6$ and added form $-1$

Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values

Break the equation in $2$ factors and set each factor equal to zero, to obtain simpler equations

Solve the equation ($1$)

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $2$ from both sides of the equation

Canceling terms on both sides

Solve the equation ($2$)

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-3$ from both sides of the equation

Canceling terms on both sides

Combining all solutions, the $2$ solutions of the equation are

The valid solutions to the logarithmic equation are the ones that, when replaced in the original equation, don't result in any logarithm of negative numbers or zero, since in those cases the logarithm does not exist