Let's calculate:

$$
x=1000^{3-2 \lg 125 / 4}
$$
It is advisable to switch to a base of 10, as a logarithm with a base of 10 appears in the exponent:

$$
x=\left(10^{3}\right)^{3-2 \lg 125 / 4}=10^{9-6 \lg 125 / 4}
$$

If it is possible to write the exponent as the logarithm of a single number, then according to the definition of the logarithm, this number is the sought-after $x$. The exponent can be written as follows:

$$
9-6 \lg \frac{1000}{32}=\lg 10^{9}-\lg \left(\frac{10^{3}}{2^{5}}\right)^{6}=\lg 10^{9}+\lg \frac{2^{30}}{10^{18}}=\lg \frac{2^{30}}{10^{9}}
$$

thus $x=2^{30} / 10^{9}$.

In the decimal system

$$
x=\left(\frac{2^{10}}{10^{3}}\right)^{3}=1,024^{3}=1,073741824
$$

Zoltán Bálint (Budapest, Materials Science, 3rd year)

