<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: $$\log_{5x+9}(x^2+6x+9)+\log_{x+3}(5x^2+24x+27)=4$$
$$\log_{5x+9}(x+3)^2+\log_{x+3}\left[(5x+9)(x+3)\right]=4$$
$$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)+\log_{x+3}(x+3)=4$$
$$2\frac{1}{\log_{x+3}(5x+9)}+\log_{x+3}(5x+9)+1=4$$
Now, substitute $u=\log_{x+3}(5x+9)$ and solve the quadratic equation for $u$.
Note that we need to assume $x+3>0$ and $5x+9>0$.
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