Logarithms

Concept:Use properties of logarithms, factorisation, and change of base to reduce the equation to a quadratic form.Explanation:First, rewrite the given equation:$\log_{(x+3)}(6x^2+28x+30) = 5 - 2\log_{(6x+10)}(x^2+6x+9)$.Factor the expressions:$6x^2+28x+30 = 2(x+3)(3x+5)$$x^2+6x+9 = (x+3)^2$.Apply $\log$ properties:$\log_{(x+3)}2(x+3)(3x+5) + 2\log_{(6x+10)}(x+3)^2 = 5$.Simplify:$\log_{(x+3)}(x+3) + \log_{(x+3)}(6x+10) + 4\log_{(6x+10)}(x+3) = 5$.Since $\log_{(x+3)}(x+3)=1$, we get:$\log_{(x+3)}(6x+10) + 4\log_{(6x+10)}(x+3) = 4$.Let $t = \log_{(x+3)}(6x+10)$. Then $\log_{(6x+10)}(x+3) = \frac{1}{t}$.The equation becomes:$t + \frac{4}{t} = 4 \implies t^2 - 4t + 4 = 0 \implies (t-2)^2 = 0 \implies t = 2$.Thus $\log_{(x+3)}(6x+10) = 2$, so $(x+3)^2 = 6x+10$.Expand: $x^2+6x+9 = 6x+10 \implies x^2 = 1 \implies x = \pm 1$.Check base conditions: $x+3>0$, $x+3\neq 1$, $6x+10>0$, $6x+10\neq 1$.Both $x=1$ and $x=-1$ satisfy all conditions and the original equation.Sum of real solutions: $1 + (-1) = 0$.Answer:$0$ (Option C).