The equation $6^{-1} = \frac{1}{6}$ can be rewritten as a logarithmic equation by applying the definition of logarithms.

The general form of a logarithmic equation is:

$b^y = x \implies \log_b(x) = y$

In your case, $6^{-1} = \frac{1}{6}$, so we can express this as:

$\log_6 \left( \frac{1}{6} \right) = -1$

Thus, the logarithmic form of the equation is:

$\log_6 \left( \frac{1}{6} \right) = -1$

Would you like further clarification on logarithms or another example?

Here are some follow-up questions:

1. How do we evaluate logarithmic expressions like $\log_6 \left( \frac{1}{6} \right)$?
2. What are some properties of logarithms that can help solve logarithmic equations?
3. How would you solve the equation $\log_6(x) = -1$?
4. Can you convert other exponential equations into logarithmic form?
5. What is the inverse relationship between logarithmic and exponential functions?

Tip: Remember, the logarithmic function is the inverse of the exponential function. This means $b^y = x$ can be rewritten as $\log_b(x) = y$, which often simplifies solving problems.