To solve the equation \(2^{x+2} + 7 = \frac{1}{2^{x-1}}\), let's break it down step by step.

### Step 1: Rewrite the equation with powers of 2.
Start with the given equation:

\[
2^{x+2} + 7 = \frac{1}{2^{x-1}}
\]

Rewrite the right-hand side:

\[
\frac{1}{2^{x-1}} = 2^{-(x-1)} = 2^{-x+1}
\]

Thus, the equation becomes:

\[
2^{x+2} + 7 = 2^{1-x}
\]

### Step 2: Isolate the powers of 2.
Move all terms involving powers of 2 to one side of the equation:

\[
2^{x+2} - 2^{1-x} = -7
\]

### Step 3: Explore possible solutions.
This equation is a bit tricky to solve directly using algebraic techniques, so we will check for possible solutions by plugging in values for \(x\).

Let's try some integer values of \(x\) and see if they satisfy the equation.

1. **Try \(x = 0\):**

   \[
   2^{0+2} + 7 = \frac{1}{2^{0-1}} \quad \Rightarrow \quad 4 + 7 = 2
   \]

   This does **not** satisfy the equation.

2. **Try \(x = 1\):**

   \[
   2^{1+2} + 7 = \frac{1}{2^{1-1}} \quad \Rightarrow \quad 8 + 7 = 1
   \]

   This does **not** satisfy the equation.

3. **Try \(x = -1\):**

   \[
   2^{-1+2} + 7 = \frac{1}{2^{-1-1}} \quad \Rightarrow \quad 2 + 7 = 4
   \]

   This does **not** satisfy the equation.

4. **Try \(x = -2\):**

   \[
   2^{-2+2} + 7 = \frac{1}{2^{-2-1}} \quad \Rightarrow \quad 2^0 + 7 = \frac{1}{2^{-3}} \quad \Rightarrow \quad 1 + 7 = 8
   \]

   This **does** satisfy the equation.

So, \(x = -2\) is a solution to the equation.

### Final Answer:
The solution to the equation \(2^{x+2} + 7 = \frac{1}{2^{x-1}}\) is:

\[
x = -2
\]

Would you like further details on how this was solved or any other steps?

### Relative Questions:
1. How can you solve equations involving powers of 2 in general?
2. What are some alternative methods for solving exponential equations besides trying specific values?
3. How does the negative exponent rule apply when solving equations like this?
4. Can you find any other solutions for this equation, or is \(x = -2\) the only one?
5. How would you solve an equation like \(3^{x+1} + 5 = \frac{1}{3^{x-2}}\)?

### Tip:
When solving exponential equations, if algebraic manipulation seems challenging, consider testing integer values or using logarithms to simplify the powers of the base.

Learn how to solve the exponential equation 2^(x+2) + 7 = 1/2^(x-1) step-by-step with detailed solutions and examples. Explore the algebraic manipulation of powers of 2 and find the solution to this challenging equation. Suitable for Grades 9-12.

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