Solve for $x$ in the equation
\[2^{(16^x)} = 16^{(2^x)}.\]
We can write
\[16^{(2^x)} = (2^4)^{(2^x)} = 2^{4 \cdot 2^x}.\]Then $2^{16^x} = 2^{4 \cdot 2^x},$ so
\[16^x = 4 \cdot 2^x.\]In turn, we can write this as
\[2^{4x} = 2^{x + 2},\]so $4x = x + 2.$  Therefore, $x = \boxed{\frac{2}{3}}.$
\frac{2}{3}
Level 2
Intermediate Algebra
Problem: Solve for $x$ in the equation
\[2^{(16^x)} = 16^{(2^x)}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation \(2^{(16^x)} = 16^{(2^x)}\), we first express both sides in terms of base 2. We know that \(16\) can be rewritten as \(2^4\). Therefore, we can substitute \(16\) in the equation:

The left side becomes:
\[ 
2^{(16^x)} = 2^{((2^4)^x)} = 2^{(2^{4x})} 
\]

The right side becomes:
\[ 
16^{(2^x)} = (2^4)^{(2^x)} = 2^{(4 \cdot 2^x)} 
\]

Now we have the equation:
\[ 
2^{(2^{4x})} = 2^{(4 \cdot 2^x)} 
\]

Since the bases are the same, we can equate the exponents:
\[ 
2^{4x} = 4 \cdot 2^x 
\]

Next, we can express \(4\) as \(2^2\):
\[ 
2^{4x} = 2^2 \cdot 2^x = 2^{(2 + x)} 
\]

Now we equate the exponents again:
\[ 
4x = 2 + x 
\]

To isolate \(x\), we subtract \(x\) from both sides:
\[ 
4x - x = 2 
\]
\[ 
3x = 2 
\]

Dividing both sides by 3 gives:
\[ 
x = \frac{2}{3} 
\]

Thus, the final answer is \(\frac{2}{3}\).<|END_OF_LATENT|> The final answer is $\frac{2}{3}$.