<s>[INST] 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?
 [/INST] HINT.-
$$\log_{5x+9}(x+3)^2=X\iff (5x+9)^X=(x+3)^2\\
\log_{x+3}(5x+9)=Y\iff(x+3)^Y=5x+9$$ It follows $(x+3)^{XY}=(x+3)^2$.
Consequently $X+Y=3$ and $XY=2$ which produces $X^2-3X+2=(X-2)(X-1)=0$ From which the solution.
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