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 \( \log_b \left( \dfrac{1}{a} \right) = -\log_b(a) \).

Learn how to evaluate the logarithmic expression \log_6\left( \dfrac{1}{36} \right) with a detailed step-by-step solution. This problem covers key concepts such as logarithms, exponents, and the logarithmic exponent rule. Suitable for students in Grades 9-11.

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