The given exponential equation is:

$8^{-2} = \frac{1}{64}$

To rewrite this in logarithmic form, we can use the relationship between logarithms and exponents. The general form is:

$a^b = c \quad \text{is equivalent to} \quad \log_a(c) = b$

In this case, the base $a = 8$, the exponent $b = -2$, and the result $c = \frac{1}{64}$.

So, the logarithmic form of the given equation is:

$\log_8\left(\frac{1}{64}\right) = -2$

Would you like further details or explanations? Let me know!

Here are 5 related questions for you:

1. What is the relationship between exponential and logarithmic functions?
2. How do you solve logarithmic equations?
3. What are the properties of logarithms?
4. How can you express a fraction like $\frac{1}{64}$ as a power of 8?
5. What does $\log_b(c) = x$ mean in terms of exponents?

Tip: When converting from exponential to logarithmic form, remember that the base of the exponential equation becomes the base of the logarithm!