<s>[INST] Solving $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$ I need help solving this logarithm exercise:

$$15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1$$

What I've done is re-writing the equation
$$\Rightarrow \qquad 5^{\log_5(3)}\cdot 3^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1 \tag{1}$$
Then applying logarithms on both sides
$$\Rightarrow \qquad \log_5(3^{\log_5(3)}\cdot5^{\log_5(3)}\cdot x^{log_5(9x)+1})=\log_5(1) \tag{2}$$
re-writing the equation a little bit
$$\Rightarrow \qquad \log_53+\log_53^{\log_5(3)}+\log_5x^{\log_5(9x)+1}=\log_55 \tag{3}$$
But then I'm not entirely sure how to proceed
 [/INST] $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1\\
x^{\log_5(9x)+1}=15^{-\log_5(3)}$
Take the $\log_5$ of both sides
$(\log_5(9x)+1)\log_5 x =(-\log_5(3))(\log_5 15)\\
(\log_5 x + \log_5 9+1)\log_5 x =(-\log_5(3))(\log_5 15)\\
(\log_5 x)^2 + (\log_5 9+1)\log_5 x + \log_5(3)(\log_5 3 + 1) = 0\\
$
let $u = \log_5 x, b = 2\log_5 3 + 1, c = (\log_5 3)^2 + \log_5 3$
$u^2 + bu + c = 0$
$u = \frac {-b \pm \sqrt {b^2 - 4c}}{2}$
$b^2 - 4c = (2\log 3 + 1)^2 - 4(\log^2 3 + \log 3)\\
4\log^2 3 + 4\log 3 + 1 - 4\log^2 3 - 4\log 3 = 1$
$u = -\log_5 3, -\log_5 3 - 1\\
x = 5^u$
$x = \frac {1}{3}$ or $\frac {1}{15}$
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