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Let's go through each part of the problem step by step.

### (a) Rewrite as an exponential equation.
We are given the logarithmic equation:
\[
\log_8 \left(\frac{1}{64}\right) = -2
\]
The general form of a logarithmic equation \(\log_b(x) = y\) is equivalent to the exponential form \(b^y = x\). Using this:

- Base \(b = 8\)
- Exponent \(y = -2\)
- Result \(x = \frac{1}{64}\)

So, converting \(\log_8 \left(\frac{1}{64}\right) = -2\) to its exponential form gives:
\[
8^{-2} = \frac{1}{64}
\]

### (b) Rewrite as a logarithmic equation.
We are given the exponential equation:
\[
3^0 = 1
\]
The general form of an exponential equation \(b^y = x\) is equivalent to the logarithmic form \(\log_b(x) = y\). Using this:

- Base \(b = 3\)
- Exponent \(y = 0\)
- Result \(x = 1\)

So, converting \(3^0 = 1\) to its logarithmic form gives:
\[
\log_3(1) = 0
\]

---

### Final Answers:
- **(a)** \(8^{-2} = \frac{1}{64}\)
- **(b)** \(\log_3(1) = 0\)

Would you like further clarification on these steps or need more details?

Here are 5 related questions to expand on this topic:
1. How can you generalize the process of converting logarithmic to exponential form?
2. Why does any number raised to the power of zero equal one?
3. Can you convert \(\log_b(1)\) for any base \(b\) to an exponential form?
4. What is the inverse relationship between exponents and logarithms?
5. How would you solve a logarithmic equation where the result is a fraction?

### Tip:
To solve logarithmic equations, remember the basic property: \(\log_b(x) = y\) means \(b^y = x\). This relationship helps in switching between exponential and logarithmic forms.

What is the answer to the problem: (a) Rewrite as an exponential equation log_8(1/64) = -2, and (b) Rewrite as a logarithmic equation 3^0 = 1?

Learn how to convert between logarithmic and exponential forms. This problem covers converting log_8(1/64) = -2 to 8^-2 = 1/64 and rewriting 3^0 = 1 as log_3(1) = 0. Ideal for high school students studying logarithms and exponents.

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