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.

Learn how to solve the exponential equation 4^x = (2^x)^2 using exponent rules. This problem involves expressing both sides with the same base and simplifying the equation. Suitable for students in Grades 8-10 studying exponential functions and algebra.

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