Task A-3.1. (8 points)

Solve the equation $\quad 4^{\log x}-32+x^{\log 4}=0$.
## Solution.

Since $4^{\log x}=x^{\log 4}$,

the equation is equivalent to

$$
2 \cdot 4^{\log x}=32, \quad 4^{\log x}=16, \quad \log x=2
$$

and finally $\quad x=100$.

For 4 points: at least (*)

For 8 points: correct result and procedure.