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 $x$ for which the given equation holds, we start with the equation itself:
${\log }_{x-3}(x^2-9)={\log }_{x-3}(x+1)+2$

Step 1: Determine the domain restrictions
For a logarithm ${\log }_b\left(a\right)$ to be defined on real numbers, the base $b$ must be positive and not equal to 1, and the argument $a$ must be positive.
Applying these rules to our equation, we get the following conditions:
1) For the base:
$x-3>0\Rightarrow x>3$
and
$x-3\ne 1\Rightarrow x\ne 4$
2) For the arguments of the logarithms:
$x^2-9>0\Rightarrow (x-3)(x+3)>0$
Since $x>3$, this condition is automatically satisfied.
3) For the second argument:
$x+1>0\Rightarrow x>-1$
Combining all these constraints, the permissible domain for $x$ is:
$x>3\text{ and }x\ne 4$

Step 2: Solve the logarithmic equation
We rewrite the constant term 2 as a logarithm with base $x-3$:
$2={\log }_{x-3}{(x-3)}^2$
Substitute this back into the original equation:
${\log }_{x-3}(x^2-9)={\log }_{x-3}(x+1)+{\log }_{x-3}{(x-3)}^2$

Apply the product rule for logarithms on the right side, ${\log }_b\left(u\right)+{\log }_b\left(v\right)={\log }_b(u\cdot v)$:
${\log }_{x-3}(x^2-9)={\log }_{x-3}\left[(x+1){(x-3)}^2\right]$

Since the logarithmic functions on both sides have the same base, we can equate their arguments:
$x^2-9=(x+1){(x-3)}^2$

Factor the left side using the difference of squares, $x^2-9=(x-3)(x+3)$:
$(x-3)(x+3)=(x+1){(x-3)}^2$

Since we established that $x>3$, we know that $x-3\ne 0$. Therefore, we can safely divide both sides of the equation by $(x-3)$:
$x+3=(x+1)(x-3)$

Expand the right side:
$x+3=x^2-3x+x-3$
$x+3=x^2-2x-3$

Rearrange the terms into a standard quadratic equation form $a{x}^2+bx+c=0$:
$x^2-3x-6=0$

Step 3: Solve the quadratic equation
Using the quadratic formula, $x=\frac{-b\cdot \pm \sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-3$, and $c=-6$:
$x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4\left(1\right)(-6)}}{2\left(1\right)}$
$x=\frac{3\pm \sqrt{9+24}}{2}$
$x=\frac{3\pm \sqrt{33}}{2}$

This gives us two potential solutions:
1) $x_1=\frac{3+\sqrt{33}}{2}$
Since $5<\sqrt{33}<6$, we have:
$\frac{3+5}{2}<x_1<\frac{3+6}{2}\Rightarrow 4<x_1<4.5$
Because $x_1>3$ and $x_1\ne 4$, this value lies within our defined domain.

2) $x_2=\frac{3-\sqrt{33}}{2}$
Since $5<\sqrt{33}<6$, we have:
$\frac{3-6}{2}<x_2<\frac{3-5}{2}\Rightarrow -1.5<x_2<-1$
Because $x_2<3$, this value is outside the domain of the logarithm and must be discarded.

Therefore, the only valid real solution for $x$ is:
$x=\frac{3+\sqrt{33}}{2}$

Since there is only one valid real value of $x$, the sum of all possible real values of $x$ is simply this single value:
$\sum x=\frac{3+\sqrt{33}}{2}$