Final answer to the problem
Step-by-step Solution
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Apply the formula: $a\log_{b}\left(x\right)$$=\log_{b}\left(x^a\right)$
Apply the formula: $\log_{b}\left(x\right)-\log_{b}\left(y\right)$$=\log_{b}\left(\frac{x}{y}\right)$, where $b=10$, $x=x^2$ and $y=x+6$
Apply the formula: $\log_{b}\left(x\right)=a$$\to \log_{b}\left(x\right)=\log_{b}\left(b^a\right)$, where $a=0$, $b=10$, $x=\frac{x^2}{x+6}$ and $b,x=10,\frac{x^2}{x+6}$
Apply the formula: $\log_{a}\left(x\right)=\log_{a}\left(y\right)$$\to x=y$, where $a=10$, $x=\frac{x^2}{x+6}$ and $y=1$
Apply the formula: $\frac{a}{b}=c$$\to a=cb$, where $a=x^2$, $b=x+6$ and $c=1$
Move everything to the left hand side of the equation
Factor the trinomial $x^2-x-6$ finding two numbers that multiply to form $-6$ and added form $-1$
Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values
Break the equation in $2$ factors and set each factor equal to zero, to obtain simpler equations
Solve the equation ($1$)
Apply the formula: $x+a=b$$\to x+a-a=b-a$, where $a=2$, $b=0$, $x+a=b=x+2=0$ and $x+a=x+2$
Apply the formula: $x+a+c=b+f$$\to x=b-a$, where $a=2$, $b=0$, $c=-2$ and $f=-2$
Solve the equation ($2$)
Apply the formula: $x+a=b$$\to x+a-a=b-a$, where $a=-3$, $b=0$, $x+a=b=x-3=0$ and $x+a=x-3$
Apply the formula: $x+a+c=b+f$$\to x=b-a$, where $a=-3$, $b=0$, $c=3$ and $f=3$
Combining all solutions, the $2$ solutions of the equation are
Verify that the solutions obtained are valid in the initial equation
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
