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
 A: Let $t=\log_5x$ and $a=\log_53$ to rewrite the given equation as
$$t^2+(1+2a)t + (a+a^2) = 0$$
which factorizes $(t+a)(t+1+a)=0$
and yields $t = -a$ and $t=-1-a$, hence $$x=\frac13,\>\frac1{15}$$
A: I use the change of base to get in terms of $e$ and $\ln$ and make this much easier.
$$\log_a b=\frac{\ln b}{\ln a} \text{ and } p^q=e^{q\ln p}$$
We get: $\log_5 3=\frac{\ln 3}{\ln 5}$, $\log_5(9x+1)=\frac{\ln(9x+1)}{\ln 5}$ and reduce our equation to:
$$e^{\frac{\ln(15)\ln(3)+\ln(x)\ln(9x+1)}{\ln 5}}=1$$
So we solve:
$$\frac{\ln(15)\ln(3)+\ln(x)\ln(9x+1)}{\ln 5}=0$$
A: I doubt that there are solutions. Observe that if $x=15,$ then necessarily we should have $\log_5(9x)+1=-\log_53.$ But these are inconsistent.
A: That last edit makes it possible. $\log_5(9x)+1=\log_5(3)+\log_5(3)+\log_5(x)+\log_5(5)=\log_5(x)+\log_5(3)+\log_5(15)$. then on takings logarithms base $5$,
$$\begin{align}\left(\log_5(15)\right)\left(\log_5(3)\right)+\left(\log_5(x)\right)\left(\log_5(x)+\log_5(3)+\log_5(15)\right)\\
=\left((\log_5(x)+\log_5(3)\right)\left(\log_5(x)+\log_5(15)\right)=0\end{align}$$
Thus either $\log_5(x)=-\log_5(3)$, so $x=1/3$, or $\log_5(x)=-\log_5(15)$ and $x=1/15$.
A: First, note that $\log_5(9x)+1 = \log_5(9x)+\log_5(5) = \log_5(45x)$:
$$15^{\log_5(3)}\cdot x^{\log_5(9x)+1} = 1 \iff 15^{\log_5(3)}\cdot x^{\log_5(45x)} = 1$$
Now, isolating the factor with $x$'s, you get
$$x^{\log_5(45x)} = \frac{1}{15^{\log_5(3)}} = 15^{-\log_5(3)} = 15^{\log_5\left(\frac{1}{3}\right)}$$
$$x^{\log_5(45x)} = 15^{\log_5\left(\frac{1}{3}\right)}$$
The first solution is found by comparing both sides. This gives $x = \dfrac{1}{15}$.
For the second solution, notice that $a^{\log_a(c)} = c$, so by changing bases, $a^{\frac{\log_b(c)}{\log_b(a)}} = c \iff a^{\log_b(c)} = c^{\log_b(a)}$. This means that $15^{\log_5\left(\frac{1}{3}\right)} = \left(\frac{1}{3}\right)^{\log_5(15)}$. Hence,
$$x^{\log_5(45x)} = \left(\frac{1}{3}\right)^{\log_5(15)}$$
This gives $x = \dfrac{1}{3}$. You could've also re-written the LHS rather than the RHS, but that would give the same result anyway.
A: $15^{\log_5(3)}\cdot x^{\log_5(9x)+1}=1\\
x^{\log_5(9x)+1}=15^{-\log_5(3)}$
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 $\frac {1}{15}$
