Trying to solve $2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$ without seeing the obvious solution 
Determine all real values of $x$ such that $$\log_{5x+9}(x^2+6x+9)+\log_{x+3}(5x^2+24x+27)=4$$

Taken from Waterloo 2012: Link
What I tried:
$$2\log_{5x+9}(x+3)+\log_{x+3}\left((x+3)(5x+9)\right)=4$$
$$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$$
Then I was stuck and I graphed it:

It was clearly $0$:
$$2\log_{9}(3)+\log_{3}(9)=3$$
But how could I do it without seeing the "zero"? Is there a way?

To complete @MrYouMath
$2u+\frac{1}{u} = 3$
where $u = \log_{5x+9} x+3$
$u = 1$ or $u = \frac{1}{2}$
Case 1: $u = 1$
$x+3 = 5x + 9$
$x = -\frac{3}{2}$
Case 2 : $u = \frac{1}{2}$
$(x+3)^2 = 5x+9$
$ x^2+x = 0$
$x = 0$ or $x= -1$
All three values are greater than $x = -3$ and $x = -\frac{9}{5}$
