The sum of all the real solutions of the equa | vedclass

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(C) Given equation: $ \log_{(x+3)}(6x^{2}+28x+30)=5-2\log_{(6x+10)}(x+3)^{2} $Factorize the argument: $ 6x^{2}+28x+30 = 2(3x^{2}+14x+15) = 2(3x+5)(x+3

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(C) Given equation: $ \log_{(x+3)}(6x^{2}+28x+30)=5-2\log_{(6x+10)}(x+3)^{2} $Factorize the argument: $ 6x^{2}+28x+30 = 2(3x^{2}+14x+15) = 2(3x+5)(x+3) $. This does not simplify directly. Let us re-examine: $ 6x^{2}+28x+30 = (6x+10)(x+3) $.So, $ \log_{(x+3)}[(6x+10)(x+3)] = 5 - 4\log_{(6x+10)}(x+3) $$ 1 + \log_{(x+3)}(6x+10) = 5 - 4\log_{(6x+10)}(x+3) $Let $ A = \log_{(x+3)}(6x+10) $. Then $ \log_{(6x+10)}(x+3) = \frac{1}{A} $.The equation becomes $ 1 + A = 5 - \frac{4}{A} $.$ A - 4 + \frac{4}{A} = 0 \Rightarrow A^{2} - 4A + 4 = 0 $.$ (A-2)^{2} = 0 \Rightarrow A = 2 $.$ \log_{(x+3)}(6x+10) = 2 \Rightarrow 6x+10 = (x+3)^{2} $.$ 6x+10 = x^{2}+6x+9 \Rightarrow x^{2} = 1 $.$ x = 1 $ or $ x = -1 $.Check domain: For $ x=1 $, base $ x+3=4 > 0, 
eq 1 $ and argument $ 6(1)^{2}+28(1)+30 = 64 > 0 $. Valid.For $ x=-1 $, base $ x+3=2 > 0, 
eq 1 $ and argument $ 6(-1)^{2}+28(-1)+30 = 8 > 0 $. Valid.Sum of solutions $= 1 + (-1) = 0$.

The sum of all the real solutions of the equation $ \log_{(x+3)}(6x^{2}+28x+30)=5-2\log_{(6x+10)}(x^{2}+6x+9) $ is equal to:

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