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:

Question

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:

Accepted Answer

C. Factor the expressions:
$$
6𝑥^2+28𝑥+30=2(𝑥+3)(3𝑥+5),
$$
$$
𝑥^2+6𝑥+9=(𝑥+3)^2.
$$

Using logarithm properties,
$$
\log_{(𝑥+3)}2+\log_{(𝑥+3)}(𝑥+3)+\log_{(𝑥+3)}(3𝑥+5)+2\log_{(6𝑥+10)}(𝑥+3)=5.
$$

Since
$$
\log_{(𝑥+3)}2+1=\log_{(𝑥+3)}(6𝑥+10),
$$
the equation becomes
$$
\log_{(𝑥+3)}(6𝑥+10)+4\log_{(6𝑥+10)}(𝑥+3)=4.
$$

Let
$$
t=\log_{(𝑥+3)}(6𝑥+10).
$$

Then
$$
\log_{(6𝑥+10)}(𝑥+3)=\frac1t,
$$
so
$$
t+\frac4t=4
\Rightarrow
t^2-4t+4=0
\Rightarrow
(t-2)^2=0.
$$

Hence,
$$
t=2.
$$

Checking the original equation gives the valid solutions
$$
𝑥=1,\,-1.
$$

Therefore,
$$
1+(-1)=0.
$$

The sum of all the real solutions of the equation log⁡(𝑥+3)(6𝑥2+28𝑥+30)=5−2log⁡(6𝑥+10)(𝑥2+6𝑥+9)\log_{(𝑥+3)}(6𝑥^2+28𝑥+30)=5-2\log_{(6𝑥+10)}(𝑥^2+6𝑥+9)log(x+3)​(6x2+28x+30)=5−2log(6x+10)​(x2+6x+9) is equal to: