$7.333 \sqrt{\log _{2}\left(2 x^{2}\right) \cdot \log _{4}(16 x)}=\log _{4} x^{3}$.
## Solution.

$$
\text { So, } \sqrt{\left(1+2 \log _{2} x\right)\left(2+\frac{1}{2} \log _{2} x\right)}=\frac{3}{2} \log _{2} x
$$

Let $y=\log _{2} x$, then $\sqrt{(2+4 y)(4+y)}=3 y \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}y \geq 0, \\ (2+4 y)(y+4)=9 y^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}y \geq 0, \\ 5 y^{2}-18 y-8=0\end{array} \Leftrightarrow y=4 \Leftrightarrow \log _{2} x=4 . \Leftrightarrow x=16\right.\right.$.

Answer: $x=16$.

Solve the systems of equations (7.334-7.340):

$7.334\left\{\begin{array}{l}\log _{2}(u+v)-\log _{3}(u-v)=1, \\ u^{2}-v^{2}=2 .\end{array}\right.$

Solution. Logarithmize the second equation:

$\left\{\begin{array}{l}\log _{2}(u+v)-\log _{3}(u-v)=1, \\ \log _{2}(u+v)+\log _{2}(u-v)=1\end{array} \Leftrightarrow\left\{\begin{array}{l}-\log _{3}(u-v)=\log _{2}(u-v), \\ (u-v)(u+v)=2, \\ u+v>0\end{array} \Leftrightarrow\right.\right.$

$$
\Leftrightarrow\left\{\begin{array} { l } 
{ u + v > 0 , } \\
{ - \frac { \log _ { 2 } ( u - v ) } { \log _ { 2 } 3 } = \log _ { 2 } ( u - v ) , } \\
{ ( u - v ) ( u + v ) = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \log _ { 2 } ( u - v ) = 0 , } \\
{ ( u - v ) ( u + v ) = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ u - v = 1 , } \\
{ u + v = 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u=\frac{3}{2} \\
v=\frac{1}{2}
\end{array}\right.\right.\right.\right.
$$

Answer: $\left(\frac{3}{2} ; \frac{1}{2}\right)$.

$7.335\left\{\begin{array}{l}x^{p}=y^{q} \\ \log _{a} \frac{x}{y}=\frac{\log _{a} x}{\log _{a} y}, p \neq q, p q \neq 0 .\end{array}\right.$

Solution. Logarithmize the first equation:

$\left\{\begin{array}{l}p \log _{a} x=q \log _{a} y, \\ \log _{a} x-\log _{a} y=\frac{\log _{a} x}{\log _{a} y}\end{array} \Leftrightarrow\left\{\begin{array}{l}p \log _{a} x=q \log _{a} y, \\ \frac{q}{p} \log _{a} y-\log _{a} y=\frac{q}{p}\end{array} \Rightarrow\left\{\begin{array}{l}\log _{a} y=\frac{q}{q-p}, \\ \log _{a} x=\frac{q^{2}}{p(q-p)}\end{array} \Rightarrow\right.\right.\right.$ $\Rightarrow\left\{\begin{array}{l}y=a^{\frac{q}{q-p}} \\ x=a^{\frac{q^{2}}{p(q-p)}}\end{array}\right.$

Answer: $\left(a^{\frac{q^{2}}{p(q-p)}}, a^{\frac{q}{q-p}}\right)$, where $a>0, a \neq 1$.

$7.336\left\{\begin{array}{l}3^{\lg x}=4^{\lg y}, \\ (4 x)^{\lg 4}=(3 y)^{\lg 3} .\end{array}\right.$

Solution. Logarithmize both equations:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \lg x \cdot \lg 3 = \lg y \cdot \lg 4 , } \\
{ ( \lg 4 + \lg x ) \lg 4 = ( \lg 3 + \lg y ) \lg 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\lg x=\lg y \cdot \frac{\lg 4}{\lg 3}, \\
\left(1+\lg y \cdot \frac{1}{\lg 3}\right) \lg ^{2} 4=\left(1+\lg y \cdot \frac{1}{\lg 3}\right) \lg ^{2} 3
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \prime } \\
{ \lg x = \lg y \cdot \frac { \lg 4 } { \lg 3 } , } \\
{ \lg y + \lg 3 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \lg x = \lg \frac { 1 } { 4 } , } \\
{ \lg y = \lg \frac { 1 } { 3 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{1}{4}, \\
y=\frac{1}{3}
\end{array}\right.\right.\right.
\end{aligned}
$$

Answer: $\left(\frac{1}{4} ; \frac{1}{3}\right)$.