2B. Let $a$ and $b$ be rational numbers such that $a>1, b>0$ and $a b=a^{b}$ and $\frac{a}{b}=a^{3 b}$. Determine the values of $a$?
Solution. First method. By taking the logarithm of the given equations with base $a$, we get

$$
\begin{array}{ll}
\log _{a} a b=\log _{a} a^{b} & \log _{a} \frac{a}{b}=\log _{a} a^{3 b} \\
1+\log _{a} b=b & 1-\log _{a} b=3 b \\
\log _{a} b=b-1 & \log _{a} b=1-3 b
\end{array}
$$

Now from the last two equations, we have $b-1=1-3 b$, i.e., $b=\frac{1}{2}$. Substituting into the equation $a b=a^{b}$, we get $a \cdot \frac{1}{2}=a^{\frac{1}{2}}$, i.e., $a^{2}=4 a$. Since $a>1$, we get $a=4$.

Second method. From $\frac{a}{b}=a^{3 b}$, it follows that $b=a^{1-3 b}$. If we substitute into $a b=a^{b}$, we will get $a^{2-3 b}=a^{b}$. From the condition $a>1$, we get $2-3 b=b, b=\frac{1}{2}$. Now if we substitute into $a b=a^{b}$, we have $\frac{a}{2}=\sqrt{a}$, from which we get that the solution is $a=4$.

Thus, $a=4$ is the only number for which the equalities are satisfied.