Phương trình đã cho tương đương với:
$3\left(\dfrac{8}{27}\right)^x+4\left(\dfrac{12}{27}\right)^x-\left(\dfrac{18}{27}\right)^x-2=0$
$\Leftrightarrow 3\left(\dfrac{2}{3}\right)^{3x}+4\left(\dfrac{2}{3}\right)^{2x}-\left(\dfrac{2}{3}\right)^x-2=0$
$\Leftrightarrow \left(\dfrac{2}{3}\right)^x=\dfrac{2}{3}$
$\Leftrightarrow x=1$