The sum of all possible real values of x for which ${\log }_{\mathrm{x-3}}(x^2-9)={\log }_{\mathrm{x-3}}(x+1)+2$ ,is

The correct option is:
$\frac{3+\sqrt{33}}{2}$

To find the sum of all possible real values of x, let's solve the given logarithmic equation step-by-step:

${\log }_{x-3}(x^2-9)={\log }_{x-3}(x+1)+2$

Step 1: Determine the domain for valid real solutions
For a logarithm ${\log }_b\left(A\right)$ to be defined:
1. The base must be positive and not equal to 1:
$x-3>0\Rightarrow x>3$
$x-3\ne 1\Rightarrow x\ne 4$
2. The arguments must be positive:
$x^2-9>0$ (which is satisfied for all $x>3$)
$x+1>0\Rightarrow x>-1$ (which is also automatically satisfied for $x>3$)
Thus, any valid real solution must satisfy $x>3$ and $x\ne 4$.

Step 2: Simplify the logarithmic equation
Using the logarithmic identity $2={\log }_{x-3}(x-3{)}^2$, we rewrite the equation:

${\log }_{x-3}(x^2-9)={\log }_{x-3}(x+1)+{\log }_{x-3}(x-3{)}^2$

Using the product rule for logarithms ${\log }_b\left(A\right)+{\log }_b\left(B\right)={\log }_b(A·B)$:

${\log }_{x-3}(x^2-9)={\log }_{x-3}\left[\right(x+1\left)\right(x-3{)}^2]$

Step 3: Solve for x
Equating the arguments of the logarithms:

$x^2-9=(x+1)(x-3{)}^2$

Factor the left-hand side as a difference of squares:

$(x-3)(x+3)=(x+1)(x-3{)}^2$

Since $x>3$, we know $x-3\ne 0$. We can divide both sides by $(x-3)$:

$x+3=(x+1)(x-3)$

Expanding the right-hand side:

$x+3=x^2-2x-3$

Rearranging into standard quadratic form:

$x^2-3x-6=0$

Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

$x=\frac{3\pm \sqrt{(-3{)}^2-4(1\left)\right(-6)}}{2}=\frac{3\pm \sqrt{9+24}}{2}=\frac{3\pm \sqrt{33}}{2}$

Step 4: Check domain validity of the roots
Since $\sqrt{33}\approx 5.744$:
1. $x_1=\frac{3+\sqrt{33}}{2}\approx 4.372$, which satisfies $x>3$ and $x\ne 4$. This is a valid solution.
2. $x_2=\frac{3-\sqrt{33}}{2}\approx -1.372$, which fails the domain condition $x>3$. Thus, it is extraneous and rejected.

Since there is only one valid real solution, the sum of all possible real values of x is simply:

$\frac{3+\sqrt{33}}{2}$