Arithmetic Derivatives: Arithmetic Logarithmic Derivative Problem In Calculus, whenever we see a constant and want to take the derivative of it, it always is 0. However in Number Theory, we have something called the arithmetic derivative in which we can differentiate to get some nonzero term. 
So we can denote the arithmetic derivative the same way as in calculus, say for some $x$, we can say $x'$ to be the arithmetic derivative. Some properties of arithmetic derivatives are that:


*

*For all primes, the arithmetic derivative is $1$.

*Product Rule: $(xy)'=x'y+xy'$

*$0'=1'=0$


Now, there is also some lesser-known sub-part to the arithmetic derivative called the Arithmetic Logarithmic Derivative, $L(n)$, which is equal to $\frac{n'}{n}$. My question is that less than $100$, how many pairs of distinct positive integers (call them $a$ and $b$) does $L(a)=L(b)$?

Since $a=b$ is a trivial solution. For $a \neq b$:
$$L(a)=L(b) \\ \frac{a'}{a}=\frac{b'}{b} \\ a'b-b'a=0 \\ \frac{a'b-b'a}{b^2}=0 \\ \left(\frac{a}{b}\right)'=0$$

Now what is $\displaystyle\left(\frac{a}{b}\right)'$? 

Lemma 1:
Let prime factorization of $n$ be $\prod_{i=1}^{k} p_i^{a_i}$, where $p_i$'s are distinct prime numbers and write $n=p_1\frac{n}{p_1}$, It follows that:
    $$n'=p_1' \frac{n}{p_1}+p_1 \left(\frac{n}{p_1} \right)'=\frac{n}{p_1}+p_1 \left(\frac{n}{p_1} \right)'=\frac{n}{p_1}+p_1 \left(p_1 \frac{n}{p_1^2} \right)' \\=\frac{n}{p_1}+p_1 \left(p_1' \frac{n}{p_1^2} +p_1 \left(\frac{n}{p_1^2} \right)' \right)=2\frac{n}{p_1}+p_1^2 \left(\frac{n}{p_1^2} \right)' $$
    Continuing this procedure we get:
    $$n'=a_1 \frac{n}{p_1}+p_1^{a_1} \left(\frac{n}{p_1^{a_1}} \right)' \ \ \ \star$$
    Now $\frac{n}{p_1^{a_1}}$ is $p_1$ free and if you do the same thing for decreasing power of $p_2$ in $\frac{n}{p_1^{a_1}}$ you will get:
    $$\left(\frac{n}{p_1^{a_1}} \right)'=a_2 \frac{n}{p_1^{a_1} p_2}+p_2^{a_2} \left(\frac{n}{p_1^{a_1} p_2^{a_2}} \right)'$$
    Put this back in the $\star$:
    $$n'=a_1 \frac{n}{p_1}+a_2 \frac{n}{p_2}+p_1^{a_1} p_2^{a_2} \left(\frac{n}{p_1^{a_1} p_1^{a_1}} \right)'$$
    Which leads us to:
    $$n'=a_1 \frac{n}{p_1}+a_2 \frac{n}{p_2}+... +a_k \frac{n}{p_k}+ p_1^{a_1} p_2^{a_2} ... p_k^{a_k} \left(\frac{n}{p_1^{a_1} p_1^{a_1} ... p_k^{a_k}} \right)' \\ n'=a_1 \frac{n}{p_1}+a_2 \frac{n}{p_2}+... +a_k \frac{n}{p_k} $$
    Here, the essence of this proof is that for a natural number $a$ with prime factorization, say $\prod_{i=1}^{k} p_i^{m_i}$, then we have:
    $$a'=a \sum_{i=1}^{k} \frac{m_i}{p_i}$$

Lemma 2:
Let us first verify that the Quotient Rule applies to Arithmetic Derivatives:
Note that:
    $$a'=\left(b\cdot\frac{a}{b}\right)'=b\cdot\left(\frac{a}{b}\right)'+b'\cdot\frac{a}{b}$$
    This is by Leibniz Rule.
So,$$b\cdot\left(\frac{a}{b}\right)'=a'-b'\cdot\frac{a}{b}=\frac{a'b-b'a}{b}$$
    $$\therefore \left(\frac{a}{b} \right)'=\frac{a'b -b'a}{b^2}$$ 

Lemma 3:
If $a=\prod_{i=1}^{k} p_i^{m_i}$ and $b=\prod_{j=1}^{l} q_j^{n_j}$
    $$\left(\frac{a}{b} \right)'=\frac{a'b -b'a}{b^2} \\ =\frac{ba\sum_{i=1}^{k} \frac{m_i}{p_i} -ab \sum_{j=1}^{l} \frac{n_j}{q_j}}{b^2} \\ =\frac{a}{b} \left(\sum_{i=1}^{k} \frac{m_i}{p_i} - \sum_{j=1}^{l} \frac{n_j}{q_j} \right)$$
    Therefore, if $\prod_{i=1}^{k} p_i^{x_i}$ is a factorization of a rational number $x$ in prime powers, (where some $x_i$ may be negative) then we can write:
    $$x'=x \sum_{i=1}^{k} \frac{x_i}{p_i}$$
    Here, the essence of this proof is to show that if we have $\frac{a}{b}=\prod_{i=1}^{k} p_i^{x_i}$ then:
    $$\left(\frac{a}{b}\right)'=\frac{a}{b} \sum_{i=1}^{k} \frac{x_i}{p_i}$$


Now for this problem:
$$\left(\frac{a}{b}\right)'=\frac{a}{b} \sum_{i=1}^{k} \frac{x_i}{p_i}=0 \\ \sum_{i=1}^{k} \frac{x_i}{p_i}=0 \\ x_1 p_2 p_3 ... p_k+x_2 p_1 p_3 ... p_k+ ... + x_k p_1 p_2 ... p_k=0 \ \ \ \ \ (1) $$
Here we can conclude that $x_i=c_i p_i$ and equation $(1)$ becomes:
$$c_1+c_2+...+c_k=0$$ Therefore the solutions to the equation $\left(\frac{a}{b}\right)'=0 \ \ \ \ (2)$ are of 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 $(2)$ then $\frac{ma}{mb}$ is a solution too. So the solutions are:
$$(2^2, 3^3), (2^3, 2\times 3^3), (3\times 2^2, 3^4), (3^3, 2^2), (2\times 3^3, 2^3), (3^4, 3\times 2^2) \\ = (4,27),(8,54),(12,81),(27,4),(54,8),(81,12)$$
