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

Factor the polynomial terms: $6x^2+28x+30 = (x+3)(6x+10)$ and $x^2+6x+9 = (x+3)^2$.
Substitute into the equation:
$\log_{(x+3)}((x+3)(6x+10)) = 5 - 2\log_{(6x+10)}((x+3)^2)$.
Using log properties: $1 + \log_{(x+3)}(6x+10) = 5 - 4\log_{(6x+10)}(x+3)$.
Let $y = \log_{(x+3)}(6x+10)$. Then $\log_{(6x+10)}(x+3) = 1/y$.
$1 + y = 5 - \frac{4}{y}$.
Multiply by $y$: $y^2 + y = 5y - 4 \implies y^2 - 4y + 4 = 0$.
$(y-2)^2 = 0 \implies y=2$.
So, $\log_{(x+3)}(6x+10) = 2$.
$6x+10 = (x+3)^2 = x^2+6x+9$.
$x^2 = 1 \implies x = 1, -1$.
Check validity: For $x=1$, base is 4 (valid), argument is 64 (valid).
For $x=-1$, base is 2 (valid), argument is 8 (valid).
Sum of solutions = $1 + (-1) = 0$.