To solve the equation $4^x = (2^x)^2$, let's break it down step by step.

Step 1: Express both sides using the same base

We know that $4$ is a power of $2$, i.e. $4 = 2^2$. So, we can rewrite the left side as:

$4^x = (2^2)^x = 2^{2x}$

Thus, the equation becomes:

$2^{2x} = (2^x)^2$

Step 2: Simplify the equation

The right-hand side is $(2^x)^2$, which can be simplified as $2^{2x}$. So, the equation becomes:

$2^{2x} = 2^{2x}$

Step 3: Conclusion

Both sides of the equation are identical, which means the equation holds true for all values of $x$. Therefore, the solution to this equation is:

$x \in \mathbb{R} \quad (\text{all real numbers})$

Summary:

This equation is true for every value of $x$. There is no specific solution for $x$, as it is true for all real numbers.

Would you like further details on how this works, or have any other questions about similar problems? Here's a few more to explore:

1. What would happen if the equation was $4^x = 2^{x+2}$?
2. How would you solve an equation like $2^{x+1} = 8^x$?
3. Can you solve exponential equations with different bases? How would you handle that?
4. What is the general method to solve exponential equations like these?
5. How do logarithms play a role in solving exponential equations?

Tip: When dealing with exponential equations, it’s helpful to express all terms with the same base, whenever possible, to simplify solving.