2. If the equality holds

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

determine $\log _{a} b, \log _{a b} b, \log _{a b^{2}} b$ and $\log _{a b^{3}} b$.
Solution. For $b=1$, all logarithms are equal to 0.

For $b>0$ and $b \neq 1$, 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 we get the equation $\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+3}$, whose solution is $x=-\frac{3}{2}$. Therefore, $\log _{a} b=-\frac{2}{3}, \log _{a b} b=-2, \log _{a b^{2}} b=2$ and $\log _{a b^{3}} b=\frac{2}{3}$.