B1. Find all real numbers $x$ that satisfy the equation

$$
\log _{2}(10 x)+\log _{4}(100 x)+\log _{8}(1000 x)-2 \log _{64} x=9
$$

Write the result in the form of a simplified fraction.

$$
\text { (6 points) }
$$
B1. We transform the equation

$$
\begin{aligned}
9 & =\log _{2}(10 x)+\log _{4}(100 x)+\log _{8}(1000 x)-2 \log _{64} x \\
& =\log _{2}(10 x)+\frac{\log (100 x)}{\log 4}+\frac{\log (1000 x)}{\log 8}-\frac{2 \log x}{\log 64}= \\
& =\log _{2}(10 x)+\frac{\log (100 x)}{2 \log 2}+\frac{\log (1000 x)}{3 \log 2}-\frac{2 \log x}{6 \log 2}= \\
& =\log _{2}(10 x)+\frac{1}{2} \log _{2}(100 x)+\frac{1}{3} \log _{2}(1000 x)-\frac{1}{3} \log _{2} x= \\
& =\log _{2}(10 x)+\log _{2}(\sqrt{100 x})+\log _{2}(\sqrt[3]{1000 x})-\log _{2} \sqrt[3]{x}= \\
& =\log _{2}\left(\frac{10 x \cdot \sqrt{100 x} \cdot \sqrt[3]{1000 x}}{\sqrt[3]{x}}\right) \\
& =\log _{2}\left(1000 \sqrt{x^{3}}\right),
\end{aligned}
$$

from which it follows that $1000 \sqrt{x^{3}}=2^{9}$ or $x=\left(\frac{2^{9}}{10^{3}}\right)^{\frac{2}{3}}=\frac{16}{25}$.

Expressing the logarithms with bases 4, 8, and 64 as quotients of logarithms with base 2 ..... 1 point
Writing the equation where all logarithms have base 2 ..... 1 point
Considering that the sum of logarithms is the logarithm of the product ..... 1 point
The equation $9=\log _{2}\left(1000 \sqrt{x^{3}}\right)$ ..... 1 point
The answer $x=\frac{16}{25}$ ..... 1 point

2. Another way The left side can be transformed as follows

$$
\begin{aligned}
& \log _{2}(10 x)+\log _{4}(100 x)+\log _{8}(1000 x)-2 \log _{64} x= \\
& =\frac{\log (10 x)}{\log 2}+\frac{\log (100 x)}{\log 4}+\frac{\log (1000 x)}{\log 8}-\frac{2 \log x}{\log 64}= \\
& =\frac{\log 10+\log x}{\log 2}+\frac{\log 100+\log x}{\log 2^{2}}+\frac{\log 1000+\log x}{\log 2^{3}}-\frac{2 \log x}{\log 2^{6}}= \\
& =\frac{1+\log x}{\log 2}+\frac{2+\log x}{2 \log 2}+\frac{3+\log x}{3 \log 2}-\frac{2 \log x}{6 \log 2}= \\
& =\frac{1}{\log 2}\left((1+1+1)+\log x\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\right)\right)= \\
& =\frac{3+\frac{3}{2} \log x}{\log 2}=\frac{6+3 \log x}{2 \log 2}
\end{aligned}
$$

Thus, we have the equation

$$
\frac{6+3 \log x}{2 \log 2}=9
$$

From this, we can express $\log x=\frac{1}{3}(18 \log 2-6)=6 \log 2-2$, so

$$
x=10^{6 \log 2-2}=\frac{\left(10^{\log 2}\right)^{6}}{10^{2}}=\frac{2^{6}}{2^{2} \cdot 5^{2}}=\frac{2^{4}}{5^{2}}=\frac{16}{25}
$$

Writing the equation where all logarithms have base 10 ..... 1 point
Considering that the logarithm of a product is the sum of the logarithms ..... 1 point
dotfill 1 point
Considering $\log 4=2 \log 2, \log 8=3 \log 2$ and $\log 64=6 \log 2$ ..... 1 point
Writing equations where only $\log x$ and $\log 2$ appear ..... 1 point
The equation $\frac{6+3 \log x}{2 \log 2}=9$ ..... 1 point
The answer $x=\frac{16}{25}$ ..... 1 point