<Question>
If $2^8=4^x$, what is the value of $x$?
</Question>

<think>
<Plan_1>
Okay, let's see. The problem is asking me to find the value of x in the equation 2^8 = 4^x. Hmm, I need to figure this out step by step. First, I remember that when dealing with exponential equations, it's often helpful to express both sides with the same base if possible. That way, I can set the exponents equal to each other since if a^b = a^c, then b must equal c, right?

So, looking at the equation 2^8 = 4^x. I know that 4 is a power of 2. Specifically, 4 is 2 squared.
So maybe I can rewrite 4 as 2^2. Let me try that.

<prompt_1.1> Rewrite the equation 2^8 = 4^x using the fact that 4 is 2 squared. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
If I substitute 4 with 2^2, the equation becomes 2^8 = (2^2)^x. Now, I need to simplify the right side. When you raise a power to another power, you multiply the exponents. So (2^2)^x would be 2^(2*x). So the equation now is 2^8 = 2^(2x).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Since the bases are the same (both are base 2), the exponents must be equal.
Therefore, 8 must equal 2x. To solve for x, I can divide both sides by 2.

<prompt_2.1> Solve the equation 8 = 2x for x by dividing both sides by 2. </prompt_2.1>

Wait, let me check if that makes sense. Let me substitute x back into the original equation to verify.

<prompt_2.2> Verify the solution by substituting x = 4 back into the original equation 2^8 = 4^x. </prompt_2.2>

Alternatively, maybe I can solve this using logarithms. Let me try that method to confirm. Taking the logarithm of both sides.

<prompt_2.3> Solve the equation 2^8 = 4^x using logarithms. </prompt_2.3>

Another way to think about it is by expressing both sides in terms of base 4. Since 2 is the square root of 4, right?
So 2 is 4^(1/2). Then 2^8 would be (4^(1/2))^8 = 4^(8*(1/2)) = 4^4. Therefore, 4^4 = 4^x, so x must be 4.

Wait, let me make sure I didn't skip any steps. Let me write that again.
If 2 = 4^(1/2), then 2^8 = (4^(1/2))^8 = 4^(8/2) = 4^4. Hence, x=4.
</Plan_2>

<Execution_2>
<execution_2.1>
So x = 8 / 2. That means x = 4.
</execution_2.1>

<execution_2.2>
The original equation is 2^8 = 4^x. If x is 4, then 4^4 should equal 2^8. Calculating 4^4: 4*4=16, 16*4=64, 64*4=256. And 2^8 is 2*2=4, 4*2=8, 8*2=16, 16*2=32, 32*2=64, 64*2=128, 128*2=256. Yep, both sides equal 256.
</execution_2.2>

<execution_2.3>
So log(2^8) = log(4^x). Using the logarithm power rule, that becomes 8*log(2) = x*log(4). Then solving for x gives x = (8*log(2))/log(4). 

But since 4 is 2^2, log(4) is log(2^2) = 2*log(2). So substituting back in, x = (8*log(2))/(2*log(2)) = 8/2 = 4. Same answer. So that checks out.
</execution_2.3>
</Execution_2>

<Final_answer>
Therefore, the value of $x$ is $\boxed{4}$.
</Final_answer>
</think>