Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$. Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$.
Is there another way to solve other than this attempt?
My attempt,
$(3-\log_3x)\log _{3x}3=1$
$\frac{\log(3)\left(3-\frac{\log (x)}{\log (3)}\right)}{\log (3x)}=1$
$\log (3)\left(3-\frac{\log (x)}{\log (3)}\right)=\log (3x)$
$3\log (3)-\log(x)=\log (3x)$
$3\log (3)-\log (x)-\log(3x)=0$
$\log (27)+\log (\frac{1}{x})+\log (\frac{1}{3x})=0$
$\log \left(\frac{27}{x(3x)}\right)=0$
$\log \left(\frac{9}{x^2}\right)=0$
$\frac{9}{x^2}=1$
$9=x^2$
$x=\pm 3$
$-3$ is rejected. So the only solution is $3$

Using another approach, recall that $\log_b x=\frac{\log_c x}{\log_cb}$.  
Now, for $b=3x$ and $c=3$ we have
$$\log_{3x}3=\frac{\log_33}{\log_3 3x}=\frac{\log_33}{\log_3 3+\log_3x}=\frac{1}{1+\log_3x}$$
Let $y=\log_3 x$.  Then, the equation
$$\left(3-\log_3x\right)\log_{3x}3=1\implies\frac{3-y}{1+y}=1\implies y=1\implies x=3$$
