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

For $\log_{x-3}(x^2-9) = \log_{x-3}(x+1) + 2$ to be defined:

Base: $x - 3 > 0$ and $x - 3 \neq 1$, so $x > 3$ and $x \neq 4$

Arguments: $x^2 - 9 > 0$ and $x + 1 > 0$, both already satisfied when $x > 3$

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Since $2 = \log_{x-3}(x-3)^2$, the equation becomes:

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

$\log_{x-3}(x^2 - 9) = \log_{x-3}\Big[(x+1)(x-3)^2\Big]$

Setting the arguments equal:

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

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

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Since $x > 3$, we have $x - 3 > 0$, so dividing both sides by $(x-3)$:

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

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

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

$x = \dfrac{3 \pm \sqrt{9 + 24}}{2} = \dfrac{3 \pm \sqrt{33}}{2}$

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$\dfrac{3 + \sqrt{33}}{2} \approx 4.37 \implies x > 3$ and $x \neq 4$ — valid

$\dfrac{3 - \sqrt{33}}{2} \approx -1.37 \implies x < 3$ — not valid

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Only one value satisfies the domain, so the sum of all possible values is $\dfrac{3+\sqrt{33}}{2}$.