Stuck with logarithm; Find the $x$ value I am trying to find the value of $x$ ... but I'm absolutely stuck, some hints would be appreciated!
$$ \log_3 (6x+2) - 2\log_3 (x)=2 $$
My work so far: 
$$\begin{align*}
\log\left(\dfrac{6x+2}{x^2}\right) &= 2^3\\
6x+2 &= 8x^2\\
-8x^2+6x+2 &= 0\\
x = 1,&\quad x = \dfrac{1}{4}
\end{align*}$$
But I've tested with 1 and 1/4 and it's never equals to 2...
Where's my error ?
Thanks !
 A: It's logarithm base 3, so your second line should read:
$$\dfrac{6x+2}{x^2} = 3^2$$
Note also that you have exponentiated, so there should be no $\ln$ anymore.
A: You can't just drop the $\log_3$, so it should be $\log_3 \left( \frac{6x+2}{x^2} \right) =2$ which leads us to $6x+2=x^2 \cdot 3^2 =9x^2$.
Now you can go on the way you did: $-9 x^2+6x+2=0$ which is the case if and only if $x= \frac{1}{3} \pm \frac{\sqrt{3}}{3}$
Edit: I replaced $\ln$ by $\log_3$ in my answer and subsequently changed the results of the calculation.
A: It is when
$$
\mathrm{log}_3 (6x+2) - 2 \mathrm{log} (x) = 2 \longrightarrow \log_3 \left( \frac{ 6x+2}{x^2} \right) = 2 \longrightarrow \frac{6x+2}{x^2} = 3^2 = 9.
$$
When you get rid of logarithms you must take "3 to the both sides".
A: I don't know how you got the first line; you should have
$$\log_3\left(\frac{6x+2}{x^2}\right) = 2,$$
not the natural log and not equal to $2^3$.
Then you would conclude from this that
$$3^{2} = \frac{6x+2}{x^2}.$$
Note in particular that you get $3^2$, not $2^3$. The logarithm base $3$ is what the exponent needs to be, not what the base needs to be.
