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 !