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?
 A: To solve 
$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$
first of all we note that 
$$\log _a b=\frac{1}{\log _b a}$$
the equation becomes
$$2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9}(x+3)}=3$$
And then set $w=\log_{5x+9}(x+3)$
$2w+\dfrac{1}{w}=3\to 2w^2-3w+1=0\to w_1=\dfrac{1}{2};\;w_2=1$
If $w=1$ we have $\log_{5x+9}(x+3)=1$ which means $5x+9=x+3\to \color{red}{x=-\dfrac{3}{2}}$
If $w=\dfrac{1}{2}$ then $\log_{5x+9}(x+3)=\dfrac{1}{2}$
that is $(5x+9)^{1/2}=x+3$
$5x+9=(x+3)^2\to x^2+x=0\to \color{red}{x=0;\;x=-1}$
Let's verify these solutions in the given equation
for $x=0$ we have $2\log_{9}(3)+\log_{3}(9)=3\to 2\cdot \frac12+2=3$ true
for $x=-1$ equation becomes $2\log_{9-5}(3-1)+\log_{3-1}(9-5)=3\to 2\log_4 2+\log_2 4=3$ true
and for $x=-\dfrac{3}{2}$ we have $2\log_{9-\frac{15}{2}}\left(3-\frac32\right)+\log_{3-\frac32}\left(9-\frac{15}{2}\right)=\log_{\frac{3}{2}}\left(\frac12\right)^2+\log_{\frac12}\left(\frac{3}{2}\right)=3$
Hope this is useful
A: 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 $5x+9>0$.
A: 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 $x = -\frac{9}{5}$
A: 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.
