To solve the equation \(\left(\frac{1}{2}\right)^{x+1} = 8\), follow these steps:

1. **Rewrite 8 as a power of 2:**
   \[
   8 = 2^3
   \]
   So the equation becomes:
   \[
   \left(\frac{1}{2}\right)^{x+1} = 2^3
   \]

2. **Rewrite \(\frac{1}{2}\) as \(2^{-1}\):**
   \[
   \left(2^{-1}\right)^{x+1} = 2^3
   \]
   This simplifies to:
   \[
   2^{-(x+1)} = 2^3
   \]

3. **Since the bases are the same, set the exponents equal to each other:**
   \[
   -(x+1) = 3
   \]

4. **Solve for \(x\):**
   \[
   -x - 1 = 3
   \]
   \[
   -x = 3 + 1
   \]
   \[
   -x = 4
   \]
   \[
   x = -4
   \]

So, the solution is \(x = -4\).

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How would the solution change if the equation were \(\left(\frac{1}{3}\right)^{x+1} = 9\)?
2. Can you solve \(\left(\frac{1}{2}\right)^{2x} = 16\) for \(x\)?
3. What happens if the equation is \(\left(\frac{1}{2}\right)^{x-1} = 32\)?
4. How do you solve \(3^{2x+1} = 81\) for \(x\)?
5. Can you solve \(5^{x+2} = 125\) for \(x\)?

**Tip:** When solving exponential equations, always try to express both sides with the same base if possible.

Learn how to solve the exponential equation (1/2)^{x+1} = 8 with a detailed step-by-step solution. This problem involves understanding exponential equations and negative exponents. Suitable for high school students.

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