We are given the logarithmic equations:

$\log_{75}(6) = a, \quad \log_{105}(36) = b, \quad \log_{6}\left(\frac{7}{5}\right) = c$

We need to find the relation between $a$, $b$, and $c$.

Step 1: Express the logarithms in terms of natural logs (or base 10 logs)

Using the change of base formula, we convert all the logarithms to a common base. The change of base formula is:

$\log_b(x) = \frac{\log(x)}{\log(b)}$

Thus, we can rewrite each logarithmic expression as follows:

1. $\log_{75}(6) = a$:

$a = \frac{\log(6)}{\log(75)}$

2. $\log_{105}(36) = b$:

$b = \frac{\log(36)}{\log(105)}$

3. $\log_6\left(\frac{7}{5}\right) = c$:

$c = \frac{\log\left(\frac{7}{5}\right)}{\log(6)} = \frac{\log(7) - \log(5)}{\log(6)}$

Step 2: Find relations between $a$, $b$, and $c$

From the expressions of $a$, $b$, and $c$, we notice that all terms involve the logarithms of various numbers. To simplify the relation, we will manipulate these expressions.

Start by isolating $\log(6)$ from $a$'s equation:

$a = \frac{\log(6)}{\log(75)} \quad \Rightarrow \quad \log(6) = a \cdot \log(75)$

Substitute $\log(6)$ in $c$'s expression:

$c = \frac{\log(7) - \log(5)}{a \cdot \log(75)}$

For $b$, express $\log(36)$ as $2\log(6)$:

$b = \frac{\log(36)}{\log(105)} = \frac{2\log(6)}{\log(105)}$

Now substitute $\log(6)$ from the expression of $a$:

$b = \frac{2a \cdot \log(75)}{\log(105)}$

Final Relation:

Thus, we now have expressions for $b$ and $c$ in terms of $a$, $\log(75)$, and $\log(105)$. The precise relationship depends on the numerical values of the logs, but we now see that $b$ is proportional to $a$, and $c$ is inversely related to $a$, involving $\log(75)$, $\log(105)$, $\log(7)$, and $\log(5)$.

Would you like further details on simplifying this relationship, or solving numerically?