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Let's solve the equation \( 4^{x+1} = 2 - 7(2^x) \) using the substitution \( u = 2^x \).

1. **Rewrite \( 4^{x+1} \) in terms of \( 2^x \):**
   Since \( 4 = 2^2 \), we can rewrite \( 4^{x+1} \) as:
   \[
   4^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x + 2}
   \]

2. **Express \( 2^{2x+2} \) in terms of \( u \):**
   Since \( u = 2^x \), we have \( 2^{2x} = (2^x)^2 = u^2 \) and therefore:
   \[
   2^{2x+2} = u^2 \cdot 2^2 = 4u^2
   \]

   Now the equation becomes:
   \[
   4u^2 = 2 - 7u
   \]

3. **Rearrange into a standard quadratic form:**
   Rewrite the equation as:
   \[
   4u^2 + 7u - 2 = 0
   \]

4. **Solve the quadratic equation \( 4u^2 + 7u - 2 = 0 \):**
   Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 7 \), and \( c = -2 \):
   \[
   u = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}
   \]
   \[
   u = \frac{-7 \pm \sqrt{49 + 32}}{8}
   \]
   \[
   u = \frac{-7 \pm \sqrt{81}}{8}
   \]
   \[
   u = \frac{-7 \pm 9}{8}
   \]

   This gives two solutions:
   \[
   u = \frac{2}{8} = \frac{1}{4} \quad \text{and} \quad u = \frac{-16}{8} = -2
   \]

5. **Convert back to \( x \):**
   Recall that \( u = 2^x \), so we solve for \( x \) in each case.

   - For \( u = \frac{1}{4} \):
     \[
     2^x = \frac{1}{4} = 2^{-2} \Rightarrow x = -2
     \]

   - For \( u = -2 \):
     Since \( u = 2^x \) and \( 2^x \) is always positive, there is no solution for \( u = -2 \).

6. **Final Answer:**
   The solution is:
   \[
   x = -2
   \]

Would you like further details, or have any questions?

Here are five related questions for deeper understanding:
1. How would you solve the equation if it were \( 4^{x+1} = 3 - 7(2^x) \) instead?
2. What would happen if you substituted \( u = 2^{x+1} \) instead of \( u = 2^x \)?
3. How can you verify that \( x = -2 \) is indeed a solution to the original equation?
4. What are the steps in solving a quadratic equation, and how do they apply here?
5. How does this approach change if the equation were an inequality?

**Tip:** Always remember to check the feasibility of each solution when dealing with substitutions, especially in exponential equations where certain values may be invalid.

Use the substitution u = 2^x to solve the equation 4^(x+1) = 2 - 7(2^x).

Learn how to solve the exponential equation 4^(x+1) = 2 - 7(2^x) by using the substitution u = 2^x, transforming it into a quadratic equation. Detailed steps include expanding and solving with the quadratic formula. Suitable for students in Grades 10-12.

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