Task B-4.6. Solve the equation $3 \cdot 4^{x}+(3 x-10) \cdot 2^{x}+3-x=0$.
## Solution.

We solve the equation $3 \cdot 4^{x}+(3 x-10) \cdot 2^{x}+3-x=0$ by substituting $2^{x}=t$. Then we have $3 \cdot t^{2}+(3 x-10) \cdot t+3-x=0$

$$
\begin{gathered}
t_{1,2}=\frac{-3 x+10 \pm \sqrt{9 x^{2}-48 x+64}}{6}=\frac{-3 x+10 \pm(3 x-8)}{6} \\
t_{1}=\frac{1}{3}, \quad t_{2}=\frac{-6 x+18}{6}=-x+3
\end{gathered}
$$

Then we have $x_{1}=-\log _{2} 3$ or $2^{x}=-x+3$.

![](https://cdn.mathpix.com/cropped/2024_05_30_bde1d6b20cdcb90302c4g-22.jpg?height=680&width=699&top_left_y=877&top_left_x=239)

Sketch

We solve the equation $2^{x}=-x+3$ graphically, or the second solution is

$$
x_{2}=1\left(\text { Verification: } 2^{1}=-1+3\right)
$$

Note: If a student guessed the second solution of the equation $2^{x}=-x+3$ without drawing or explaining that it is the only solution of the equation, give one point instead of the last three points.