https://r2.erweima.ai/mathbot/image/1725897523436eah7r8yr.jpg

It seems the image's equation text wasn't extracted clearly using the current method. However, I can still solve the equation manually as it appears from the image:

\[
\left(\frac{1}{2}\right)^{2x+1} = \sqrt{\frac{2^{4x-1}}{128}}
\]

### Solution Steps:

1. **Simplify the square root expression** on the right-hand side:
   \[
   \sqrt{\frac{2^{4x-1}}{128}} = \frac{\sqrt{2^{4x-1}}}{\sqrt{128}}
   \]
   We know that \( \sqrt{128} = \sqrt{2^7} = 2^{7/2} \), so this becomes:
   \[
   \frac{2^{(4x-1)/2}}{2^{7/2}} = 2^{\frac{4x-1}{2} - \frac{7}{2}} = 2^{\frac{4x - 1 - 7}{2}} = 2^{\frac{4x - 8}{2}} = 2^{2x - 4}
   \]

   Therefore, the equation becomes:
   \[
   \left(\frac{1}{2}\right)^{2x+1} = 2^{2x-4}
   \]

2. **Express the left-hand side** using powers of 2:
   \[
   \left(\frac{1}{2}\right)^{2x+1} = 2^{-(2x+1)}
   \]

   Now the equation is:
   \[
   2^{-(2x+1)} = 2^{2x-4}
   \]

3. **Equate the exponents** since the bases are the same:
   \[
   -(2x+1) = 2x - 4
   \]

4. **Solve for \(x\)**:
   \[
   -2x - 1 = 2x - 4
   \]
   Add \(2x\) to both sides:
   \[
   -1 = 4x - 4
   \]
   Add 4 to both sides:
   \[
   3 = 4x
   \]
   Divide by 4:
   \[
   x = \frac{3}{4}
   \]

### Final Answer:
\[
x = \frac{3}{4}
\]

Let me know if you'd like further clarification or additional details on any step! Here are some related questions:

1. What if the right-hand side was a cube root instead of a square root?
2. How would the equation change if the base on the left-hand side was \( \frac{1}{3} \) instead of \( \frac{1}{2} \)?
3. What happens if we introduce logarithms to solve such an equation?
4. Can you solve this equation using a graphical method?
5. How would you approach this equation if decimals were involved?

**Tip:** Always try to simplify expressions involving roots and powers of the same base for easier manipulation.

Solve the equation (1/2)^(2x+1) = √[(2^(4x-1))/128] using laws of exponents and radicals. The problem involves simplifying powers and equating exponents, suitable for high school students studying algebraic exponents.

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