1A. 2B. Determine $\log _{a} b, \log _{a b} b, \log _{a b^{2}} b$ and $\log _{a b^{3}} b$, if

$$
\log _{a} b-\log _{a b} b=\log _{a b^{2}} b-\log _{a b^{3}} b
$$
Solution. For $b=1$, the equality is true for any $a$, because all logarithms are equal to zero. For $b>0, b \neq 1$, the given condition can be written in the form

$$
\frac{1}{\log _{b} a}-\frac{1}{\log _{b} a+1}=\frac{1}{\log _{b} a+2}-\frac{1}{\log _{b} a+3}
$$

Let $\log _{b} a=x$. Then from the last equality, we have

$$
\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+3}
$$

from which we find $x=-\frac{3}{2}$. Therefore, the required values are

$$
\log _{a} b=-\frac{2}{3}, \log _{a b} b=-2, \log _{a b^{2}} b=2, \log _{a b} 3 b=\frac{2}{3}
$$