Task B-3.2. Solve the equation $2 \cdot 8^{2014 x}-3 \cdot 4^{2014 x}-3 \cdot 2^{2014 x}+2=0$.
## Solution.

$$
2 \cdot 8^{2014 x}-3 \cdot 4^{2014 x}-3 \cdot 2^{2014 x}+2=0
$$

We introduce the substitution $2^{2014 x}=t$.

Then we have $2 t^{3}-3 t^{2}-3 t+2=0$.

We have obtained a symmetric cubic equation. After grouping, we get

$$
\begin{aligned}
& 2\left(t^{3}+1\right)-3 t(t+1)=0 \\
& 2(t+1)\left(t^{2}-t+1\right)-3 t(t+1)=0 \\
& (t+1)\left(2 t^{2}-2 t+2-3 t\right)=0 \\
& (t+1)\left(2 t^{2}-5 t+2\right)=0 \\
& t=-1 \text { or } t=2 \text { or } t=\frac{1}{2}
\end{aligned}
$$

Since $2^{2014 x}$ is always positive, the only possible solutions are $t=2$ or $t=\frac{1}{2}$.

Then we have $2^{2014 x}=2$ or $2^{2014 x}=\frac{1}{2}$, that is,

$$
2014 x=1 \text { or } 2014 x=-1
$$

The solutions are thus $x=\frac{1}{2014}$ and $x=\frac{-1}{2014}$.

(2 points)