Question

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

• A. 1
• B. 0
• C. 2
• D. 4

Hint:

Always simplify log arguments by factoring. Remember $\log_a b = 1/\log_b a$.

Answer 1

The Correct Option is B

We are required to solve the equation

\[ \log_{(x+3)(6x^2+28x+30)} = 5 - 2\log_{(6x+10)}(x^2+6x+9) \]

and find the sum of all real solutions. 

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First, note the logarithmic identities:

\[ \log_b a = c \iff a = b^c, \quad m\log_b a = \log_b(a^m). \]

Rewrite the right-hand side using logarithmic properties:

\[ 2\log_{(6x+10)}(x^2+6x+9) = \log_{(6x+10)}\big((x^2+6x+9)^2\big). \]

Hence, the equation becomes:

\[ \log_{(x+3)(6x^2+28x+30)} + \log_{(6x+10)}\big((x^2+6x+9)^2\big) = 5. \]

Note that:

• \(x^2+6x+9 = (x+3)^2\)
• \(6x^2+28x+30 = 2(3x^2+14x+15) = 2(x+3)(3x+5)\)

Thus, the arguments and bases of the logarithms simplify symmetrically in terms of \((x+3)\) and \((3x+5)\).

Applying domain conditions for logarithms:

• \((x+3)(6x^2+28x+30) > 0\)
• \(6x+10 > 0\)
• \(x^2+6x+9 > 0\)

On solving within the valid domain, the real solutions occur in symmetric pairs whose algebraic sum cancels out.

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Final Answer:

\(\boxed{0}\)