<Question>
Solve for $x$: $2^{2x} = 256^\frac{1}{2}$.
</Question>

<think>
<Plan_1>
Okay, so I need to solve the equation 2^(2x) = 256^(1/2). Let me start by understanding what each side of the equation means. On the left side, 2^(2x) is straightforward—it's 2 raised to the power of 2x. 

The right side is 256^(1/2). I remember that a fractional exponent like 1/2 means taking the square root. So, 256^(1/2) is the square root of 256. Let me confirm that.
The square root of 256 is 16 because 16*16 = 256. So, the equation simplifies to 2^(2x) = 16.

Now, I need to solve for x here. Both sides of the equation are powers of 2 and 16, but 16 is a power of 2 as well. Let me express 16 as a power of 2.

<prompt_1.1> Express 16 as a power of 2 and rewrite the equation. </prompt_1.1>
-----

Alternatively, maybe there's another approach. Let's see. The original equation is 2^(2x) = 256^(1/2). Maybe we can write both sides with the same base.

<prompt_1.2> Explore an alternative method to solve the equation by expressing both sides with the same base. </prompt_1.2>
-----

Alternatively, if there was a mistake in expressing 256 as a power of 2, but 2^8 is 256, so that's correct.
If someone thought 256 is 2^9, that would be wrong because 2^10 is 1024, so 2^8 is 256. So that's correct.
-----

Another way to check is to use logarithms. Let's try that method. Taking the natural logarithm of both sides.

<prompt_1.3> Use logarithms to verify the solution and ensure consistency with previous methods. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Since 2^4 = 16 (because 2*2=4, 4*2=8, 8*2=16). So, 16 is 2^4. Therefore, I can rewrite the equation as 2^(2x) = 2^4.
</execution_1.1>

<execution_1.2>
256 is 2^8, right? Because 2^10 is 1024, so 2^8 is 256. So, 256^(1/2) can be written as (2^8)^(1/2). Using exponent rules, when you raise a power to another power, you multiply the exponents. So, (2^8)^(1/2) becomes 2^(8*(1/2)) which is 2^4. So, 2^4 is 16, which matches what I had before. So, the equation becomes 2^(2x) = 2^4.
</execution_1.2>

<execution_1.3>
ln(2^(2x)) = ln(16). Then, using the logarithm power rule, that becomes 2x * ln(2) = ln(16). Then, solving for x: x = ln(16) / (2 * ln(2)). 

But 16 is 2^4, so ln(16) is ln(2^4) = 4*ln(2). Substituting back, x = (4*ln(2)) / (2*ln(2)) = 4/2 = 2. So that's the same answer.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, since the bases are the same and both sides are equal, their exponents must be equal as well. So, I can set the exponents equal to each other: 2x = 4. To solve for x, I just divide both sides by 2.
That gives x = 4/2 = 2. So, x = 2.

Wait, let me check if that's correct. Let me substitute x = 2 back into the original equation to verify.

<prompt_2.1> Verify the solution by substituting x = 2 back into the original equation. </prompt_2.1>
-----

But let me make sure there are no other solutions. Since the original equation is an exponential equation, sometimes there can be multiple solutions, especially if dealing with even roots or negative numbers.

<prompt_2.2> Ensure there are no other solutions by considering the properties of exponential equations. </prompt_2.2>
-----

Based on execution_1.2:
So, that's the same as before. Then, equate the exponents: 2x = 4, so x = 2.

Alternatively, maybe someone could approach this by taking the square root of both sides first. Let's see. The original equation is 2^(2x) = 256^(1/2). If we take the square root of both sides, we get (2^(2x))^(1/2) = 256^(1/2 * 1/2) = 256^(1/4).
Wait, perhaps that's complicating things. Let me try that step again. 

Taking the square root of both sides of 2^(2x) = 256^(1/2) would give us (2^(2x))^(1/2) = (256^(1/2))^(1/2). Simplifying the left side: 2^(2x * 1/2) = 2^x. The right side: 256^(1/2 * 1/2) = 256^(1/4). Then, 256^(1/4) is the fourth root of 256. Let's calculate that.
Since 4^4 is 256? Wait, 4*4=16, 16*4=64, 64*4=256. So, yes, 4^4 = 256. Therefore, the fourth root of 256 is 4. So, the equation becomes 2^x = 4.

Again, 4 is 2^2, so 2^x = 2^2, which means x = 2.
So, this method also leads to the same answer. So, regardless of whether we simplify the original equation by expressing 256 as a power of 2 first or take the square root of both sides, we end up with x = 2.

Is there another way to approach this problem? Maybe by using logarithms from the start? Let's try that. Taking the logarithm of both sides. Let's use natural log.

<prompt_2.3> Use logarithms from the start to solve the equation and verify consistency with previous results. </prompt_2.3>
-----

Let me just recap to ensure I didn't miss any steps. 

Original equation: 2^(2x) = 256^(1/2)

Express 256 as 2^8: 256^(1/2) = (2^8)^(1/2) = 2^(8*(1/2)) = 2^4.

<prompt_2.4> Recap the solution process and confirm the solution using multiple methods. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Left side: 2^(2*2) = 2^4 = 16. Right side: 256^(1/2) = sqrt(256) = 16. So, 16 = 16.
</execution_2.1>

<execution_2.2>
But in this case, since the base is 2, which is positive, and the exponent is defined for all real numbers, but since we're dealing with an equation where both sides are positive real numbers, there should be only one real solution.
</execution_2.2>

<execution_2.3>
ln(2^(2x)) = ln(256^(1/2)). Applying the power rule for logarithms: 2x * ln(2) = (1/2) * ln(256). Then, solving for x: x = [(1/2) * ln(256)] / [2 * ln(2)] = [ln(256) / (4 * ln(2))]. 

But since 256 is 2^8, ln(256) = ln(2^8) = 8 * ln(2). Substituting back in: x = [8 * ln(2) / (4 * ln(2))] = 8/4 = 2. So, again, x = 2.
</execution_2.3>

<execution_2.4>
Therefore, equation becomes 2^(2x) = 2^4. Therefore, 2x = 4. x = 2.

Alternatively, if someone forgets to express 256 as a power of 2, they might get stuck. For example, if they try to equate 2^(2x) = 16 directly. Then, they might not know how to proceed unless they express 16 as 2^4. So, that's a necessary step.

Another possible mistake could be miscalculating the square root of 256. If someone thought it was 256^(1/2) = 256, which is incorrect, but that's probably a simple mistake. Or maybe miscalculating 256^(1/2) as something else, but since 16*16 is 256, it's safe.

Alternatively, if someone tried to take logarithms with a different base, but since the base is 2, using natural log or log base 2 both work. If we use log base 2, then log2(2^(2x)) = 2x, and log2(16) = 4, so 2x = 4, x = 2. If using natural log, ln(2^(2x)) = 2x ln(2), and ln(16) = 4 ln(2), so 2x ln(2) = 4 ln(2), divide both sides by ln(2), get 2x = 4, x=2.

So, all methods lead to x=2.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.3:
So, after verifying through multiple methods—expressing both sides with the same base, taking square roots first, using logarithms—it's clear that x = 2 is the solution.
All paths lead to the same answer, so I feel confident that x = 2 is correct.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the solution to the equation $2^{2x} = 256^\frac{1}{2}$ is $x = \boxed{2}$.
</Final_answer>
</think>