## Task B-3.2.

Determine the zeros of the function $f: \mathbf{R} \rightarrow \mathbf{R}, f(x)=\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)-2 x-1$.
## Solution.

The zeros of the given function are the solutions of the equation $f(x)=0$.

Thus, we need to solve the equation $\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)-2 x-1=0$.

The equation can be written in the following form:

$$
\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)=2 x+1
$$

or

$$
\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)=\log _{2} 2^{2 x+1}
$$

Then we have

$$
\begin{gathered}
18 \cdot 4^{x}-8 \cdot 2^{x}+1=2^{2 x+1} \\
18 \cdot 4^{x}-8 \cdot 2^{x}+1=2 \cdot 4^{x} \\
16 \cdot 4^{x}-8 \cdot 2^{x}+1=0
\end{gathered}
$$

We can introduce the substitution $t=2^{x}$ and solve the corresponding quadratic equation, or notice that the last expression is equal to $\left(4 \cdot 2^{x}-1\right)^{2}=0$. From this, we conclude that

$$
\begin{aligned}
& 2^{x}=\frac{1}{4} \\
& x=-2
\end{aligned}
$$

Note: The expression $\log _{2}\left(18 \cdot 4^{x}-8 \cdot 2^{x}+1\right)=2 x+1$ could also be transformed into $18 \cdot 4^{x}-8 \cdot 2^{x}+1=2^{2 x+1}$ using the relation $\log _{a} b=c \Leftrightarrow a^{c}=b$, and this method should also be awarded one point.

Note: The domain of the function is given in the problem, so it is not necessary to determine its natural domain. For the same reason, it is not necessary to write down the condition that must be satisfied for the corresponding logarithmic equation to make sense.