$7.3113 \log _{x} 4+2 \log _{4 x} 4+3 \log _{16 x} 4=0$.

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$7.3113 \log _{x} 4+2 \log _{4 x} 4+3 \log _{16 x} 4=0$.
Solution. The original equation is equivalent to the following:

$\frac{3}{\log _{4} x}+\frac{2}{\log _{4} x+1}+\frac{3}{\log _{4} x+2}=0$. Let $y=\log _{4} x$, then

$\frac{3}{y}+\frac{2}{y+1}+\frac{3}{y+2}=0 \Leftrightarrow\left\{\begin{array}{l}3(y+1)(y+2)+2 y(y+2)+3 y(y+1)=0, \\ y \neq 0, y \neq-1, y \neq-2\end{array} \Leftrightarrow\right.$

$\Leftrightarrow 4 y^{2}+8 y+3=0 \Leftrightarrow\left[\begin{array}{l}\log _{4} x=-\frac{1}{2}, \\ \log _{4} x=-\frac{3}{2}\end{array} \Leftrightarrow\left[\begin{array}{l}x=\frac{1}{2}, \\ x=\frac{1}{8} .\end{array}\right.\right.$

Answer: $x_{1}=\frac{1}{8}, x_{2}=\frac{1}{2}$.

$7.3124^{\log _{16} x}-3^{\log _{16} x-0.5}=3^{\log _{16} x+0.5}-2^{2 \log _{16} x-1}$.

## Solution.

$2^{2 \log _{2^{4}} x} \cdot\left(1+2^{-1}\right)=3^{\log _{16} x}\left(3^{0.5}+3^{-0.5}\right) \Leftrightarrow x^{\frac{1}{2}} \cdot \frac{3}{2}=3^{\log _{16} x} \cdot \frac{4}{\sqrt{3}} \Leftrightarrow$

$\log _{2}\left(3 \sqrt{3} \cdot x^{\frac{1}{2}}\right)=\log _{2}\left(8 \cdot 3^{\log _{16} x}\right) \Leftrightarrow \frac{3}{2} \log _{2} 3+\frac{1}{2} \log _{2} x=3+\frac{1}{4} \log _{2} 3 \cdot \log _{2} x \Leftrightarrow$

$\Leftrightarrow 6\left(\log _{2} 3-2\right)=\log _{2} x \cdot\left(\log _{2} 3-2\right) \Leftrightarrow \log _{2} x=6 \Leftrightarrow x=64$.

Answer: $x=64$.