To solve the equation \(2^x + 8 = \frac{1}{4} + 2^{1-x}\), let's follow these steps:

1. Rewrite the equation for clarity:
\[2^x + 8 = \frac{1}{4} + 2^{1-x}\]

2. Express \(\frac{1}{4}\) as \(2^{-2}\):
\[2^x + 8 = 2^{-2} + 2^{1-x}\]

3. Rewrite the equation using \(2^{-2}\):
\[2^x + 8 = 2^{-2} + 2^{1-x}\]

4. Recognize that \(2^{1-x} = \frac{2}{2^x}\):
\[2^x + 8 = 2^{-2} + \frac{2}{2^x}\]

5. Combine like terms:
\[2^x + 8 = \frac{1}{4} + \frac{2}{2^x}\]

6. To eliminate the fractions, multiply through by \(4 \cdot 2^x\):
\[4 \cdot 2^x \cdot 2^x + 4 \cdot 2^x \cdot 8 = 4 \cdot 2^x \cdot \frac{1}{4} + 4 \cdot 2^x \cdot \frac{2}{2^x}\]

7. Simplify each term:
\[4 \cdot 2^{2x} + 32 \cdot 2^x = 2^x + 8\]

8. Rearrange the terms:
\[4 \cdot 2^{2x} + 32 \cdot 2^x - 2^x - 8 = 0\]

9. Simplify further:
\[4 \cdot 2^{2x} + 31 \cdot 2^x - 8 = 0\]

10. Let \(y = 2^x\):
\[4y^2 + 31y - 8 = 0\]

11. Solve the quadratic equation \(4y^2 + 31y - 8 = 0\) using the quadratic formula:
\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
where \(a = 4\), \(b = 31\), and \(c = -8\):
\[y = \frac{-31 \pm \sqrt{31^2 - 4 \cdot 4 \cdot (-8)}}{2 \cdot 4}\]
\[y = \frac{-31 \pm \sqrt{961}}{8}\]
\[y = \frac{-31 \pm 31}{8}\]

12. This gives us two solutions:
\[y = \frac{-31 + 31}{8} = 0\]
\[y = \frac{-31 - 31}{8} = -\frac{62}{8} = -7.75\]

Since \(y = 2^x\) and \(2^x\) cannot be zero or negative, we only accept the solution that makes sense:
\[2^x = \frac{1}{2}\]
\[x = -1\]

So, the solution to the equation \(2^x + 8 = \frac{1}{4} + 2^{1-x}\) is:
\[x = -1\]

Would you like more details or have any questions? 

Here are some questions you might consider asking next:
1. Can you explain how to solve another type of exponential equation?
2. How do you solve logarithmic equations?
3. What are the properties of exponents?
4. How do you solve a system of linear equations?
5. Can you explain how to factor quadratic equations?

**Tip:** When solving exponential equations, try to express all terms with the same base to simplify the equation.

Learn how to solve the exponential equation 2^x + 8 = 1/4 + 2^(1-x) step-by-step. This problem involves exponential equations and requires manipulation of exponents and terms. Suitable for students in Grades 10-12.

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