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

3 + 33 2 \dfrac{3+ \sqrt{33}}{2} 2 3 + 33 ​ ​. Since the base of the logarithm has to be greater than zero and cannot be 1 1 1 , x x x has to be greater than 3 3 3 and cannot be 4 4 4 . Also, since x^2-9&gt;0 , x x x will be greater than 3 3 3 . The equation can be rewritten as; log ⁡ x − 3 ( x 2 − 9 ) − log ⁡ x − 3 ( x + 1 ) = 2 \log_{x-3}{

The sum of all possible real values of x 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(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

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