$\log_{x+3}((x+3)(6x+10)) = 5 - 2\log_{6x+10}(x+3)^2$ $1 + \log_{x+3}(6x+10) = 5 - 4\log_{6x+10}(x+3)$ Let $\log_{x+3}(6x+10) = A$ $\Rightarrow A + \frac{4}{A} = 4 \Rightarrow A = 2$ $\Rightarrow \log_{x+3}(6x+10) = 2$ $6x + 10 = (x+3)^2$ $x^2 - 1 = 0 \Rightarrow x = \pm 1$ Sum of roots $= 0$