To solve the equation:

$\left( \frac{1}{4} \right)^x = 64$

we can proceed by rewriting both sides of the equation in terms of powers of 4 to make it easier to solve for $x$.

1. First, note that $\frac{1}{4}$ can be written as $4^{-1}$, so: $\left( \frac{1}{4} \right)^x = (4^{-1})^x = 4^{-x}$

2. Now we rewrite 64 as a power of 4. Since $64 = 4^3$, we can substitute this into the equation: $4^{-x} = 4^3$

3. Since the bases are the same, we can set the exponents equal to each other: $-x = 3$

4. Solving for $x$, we get: $x = -3$

Therefore, the solution is: $x = -3$

Would you like further details on this solution, or do you have any other questions?

Here are five related questions to expand on this problem:

1. How would the solution change if the base was $\frac{1}{2}$ instead of $\frac{1}{4}$?
2. What if the equation were $\left(\frac{1}{4}\right)^x = 16$ instead?
3. How can logarithms be used to solve this type of exponential equation?
4. What is the solution if the right side of the equation were negative, like $\left(\frac{1}{4}\right)^x = -64$?
5. How would we approach solving $\left(\frac{1}{4}\right)^x = \frac{1}{64}$?

Tip: When solving exponential equations, expressing both sides with the same base can simplify the process.