The expression $\log_3 \left(\frac{1}{81}\right)$ asks for the logarithm of $\frac{1}{81}$ with base 3.

To solve this, we can rewrite $\frac{1}{81}$ as a power of 3. Since $81 = 3^4$, we can express $\frac{1}{81}$ as:

$\frac{1}{81} = 3^{-4}$

Now the expression becomes:

$\log_3 (3^{-4})$

Using the logarithmic identity $\log_b (b^x) = x$, this simplifies to:

$\log_3 (3^{-4}) = -4$

So, the value of $\log_3 \left(\frac{1}{81}\right)$ is -4.

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Would you like a more detailed breakdown or have any questions?

Here are five related questions for further exploration:

1. What is the general rule for evaluating logarithms of powers of the base?
2. How do you express a fraction as a negative exponent?
3. How would the result change if the base were different (e.g., $\log_2 \left(\frac{1}{81}\right)$)?
4. What is the relationship between logarithms and exponents?
5. How do logarithms help in solving exponential equations?

Tip: Always express fractions or numbers as powers of the base when simplifying logarithmic expressions.