1B. Solve the equation

$$
4^{\log _{10} x}-32+x^{\log _{10} 4}=0
$$
Solution. It is clear that the equation is defined for $x>0$. For $x>0, \log _{10} x$ can be written in the form

$$
\log _{10} x=\frac{\log _{4} x}{\log _{4} 10}=\left(\log _{4} x\right)\left(\log _{10} 4\right)
$$

from which we get

$$
4^{\log _{10} x}=4^{\left(\log _{4} x\right)\left(\log _{10} 4\right)}=\left(4^{\log _{4} x}\right)^{\log _{10} 4}=x^{\log _{10} 4}
$$

Now, the equation can be written in the form

$$
\begin{aligned}
& 4^{\log _{10} x}-32+4^{\log _{10} x}=0 \\
& 2 \cdot 4^{\log _{10} x}-32=0 \\
& 4^{\log _{10} x}=4^{2} \\
& \log _{10} x=2
\end{aligned}
$$

Thus, the solution to the equation is $x=100$.