Logarithmic Arithmetic Derivative

$a=b$ is a trivial solutions. For $a \neq b$ $L(a)=L(b) \\ \frac{a'}{a}=\frac{b'}{b} \\ a'b-b'a=0 \\ \frac{a'b-b'a}{b^2}=0 \\ \left(\frac{a}{b}\right)'=0$ From solution to this problem , if $\frac{a}{b}=\prod_{i=1}^{k} p_i^{x_i}$ then $\left(\frac{a}{b}\right)'=\frac{a}{b} \sum_{i=1}^{k} \frac{x_i}{p_i}$ Now for this problem $\left(\frac{a}{b}\right)'=\frac{a}{b} \sum_{i=1}^{k} \frac{x_i}{p_i}=0 \\ \sum_{i=1}^{k} \frac{x_i}{p_i}=0 \\ x_1 p_2 p_3 ... p_k+x_2 p_1 p_3 ... p_k+ ... + x_k p_1 p_2 ... p_k=0 \ \ \ \ \star$ In the same manner of solution to above mentioned problem one can conclude that $x_i=c_i p_i$ and equation $\star$ becomes $c_1+c_2+...+c_k=0$ . Therefore the solutions to the equation $\left(\frac{a}{b}\right)'=0 \ \ \ \dagger$ are in the form $\frac{a}{b}=\prod_{i=1}^{k} (p_i^{p_i})^{c_i}$ where $\sum c_i=0$ .

For this problem since $a$ and $b$ are less than $100$ only possible values for $\frac{a}{b}$ are $\frac{a}{b}=\frac{2^2}{3^3}, \frac{3^3}{2^2}$ Now if $\frac{a}{b}$ is a solution to the equation $\dagger$ then $\frac{ma}{mb}$ is a solution too. So the solutions are $(2^2, 3^3), (2^3, 2\times 3^3), (3\times 2^2, 3^4), (3^3, 2^2), (2\times 3^3, 2^3), (3^4, 3\times 2^2)$