Task B-3.1. Solve the equation $4^{4 \log x+1}-17 \cdot 4^{3 \log x}+17 \cdot 4^{\log x}-4=0$.
## Solution.

Let's introduce the substitution $4^{\log x}=t>0, x>0$. Then the given equation becomes

$$
4 t^{4}-17 t^{3}+17 t-4=0
$$

By factoring, we get

$4\left(t^{4}-1\right)-17 t\left(t^{2}-1\right)=0$

$\left(t^{2}-1\right)\left(4 t^{2}-17 t+4\right)=0$.

(1 point)

1. $t^{2}-1=0$

$t_{1}=1$ (since $t>0$)

$4^{\log x}=1$ which gives $x_{1}=1$.

(1 point)
2. $4 t^{2}-17 t+4=0$

$t_{2}=\frac{1}{4}, t_{3}=4$

$4^{\log x}=\frac{1}{4}, 4^{\log x}=4$ and finally $x_{2}=\frac{1}{10}$ and $x_{3}=10$.

(2 points)