We are tasked with evaluating the expression $\log_{6} \left( \dfrac{1}{36} \right)$.

Step-by-step solution:

1. Let the expression be equal to $x$: $\log_{6} \left( \dfrac{1}{36} \right) = x$

2. By the definition of logarithms, we can rewrite this as an exponential equation: $6^x = \dfrac{1}{36}$

3. Notice that $36$ is a power of 6: $36 = 6^2$

   So, we substitute this into the equation: $6^x = \dfrac{1}{6^2}$

4. We know that $\dfrac{1}{6^2}$ can be written as $6^{-2}$: $6^x = 6^{-2}$

5. Since the bases are the same, we can equate the exponents: $x = -2$

Thus, $\log_{6} \left( \dfrac{1}{36} \right) = -2$.

Final answer:

$\log_{6} \left( \dfrac{1}{36} \right) = -2$

Would you like further details on logarithmic properties or have any questions? Here are some related questions for practice:

1. Evaluate $\log_{2} \left( \dfrac{1}{8} \right)$.
2. Simplify $\log_{10} \left( \dfrac{1}{100} \right)$.
3. Solve $\log_{5} \left( \dfrac{1}{25} \right)$.
4. Find $\log_{4} \left( \dfrac{1}{16} \right)$.
5. Evaluate $\log_{7} \left( \dfrac{1}{49} \right)$.

Tip: Remember that $\log_b \left( \dfrac{1}{a} \right) = -\log_b(a)$.