CAT 2025 Slot 3 QA Question 12: The sum of all possible real values of $x$ for which $\log_{x-3}(x^2 - 9) = \log_{x-3}(x+1) + 2$, is

Question

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

Accepted Answer

The base is $x-3$, so $x > 3$ and $x 
eq 4$. The arguments are then positive.
$\log_{x-3}(x^{2} - 9) = \log_{x-3}(x + 1) + 2$
$\log_{x-3}(\dfrac{x^{2} - 9}{x + 1}) = 2$
$\dfrac{x^{2} - 9}{x + 1} = (x - 3)^{2}$
Since $x > 3$, $x-3$ is non-zero. Divide by $x-3$.
$x + 3 = (x + 1)(x - 3)$
$x + 3 = x^{2

The sum of all possible real values of xxx for which log⁡x−3(x2−9)=log⁡x−3(x+1)+2\log_{x-3}(x^2 - 9) = \log_{x-3}(x+1) + 2logx−3​(x2−9)=logx−3​(x+1)+2, is

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