To solve the equation:

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

### Step 1: Rewrite \(4^{x+1}\) in terms of \(2^x\).
Since \(4 = 2^2\), we can write \(4^{x+1}\) as \((2^2)^{x+1}\), which simplifies to:
\[
4^{x+1} = 2^{2(x+1)} = 2^{2x + 2}
\]

So, the equation becomes:
\[
2^{2x + 2} = 2 - 7 \cdot 2^x
\]

### Step 2: Substitute \(y = 2^x\).
Let \(y = 2^x\), so \(2^{2x} = y^2\). This substitution changes the equation to:
\[
y^2 \cdot 4 = 2 - 7y
\]
or equivalently,
\[
4y^2 + 7y - 2 = 0
\]

### Step 3: Solve the quadratic equation.
Now we have a quadratic equation in \(y\):
\[
4y^2 + 7y - 2 = 0
\]

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4\), \(b = 7\), and \(c = -2\):

\[
y = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}
\]

\[
y = \frac{-7 \pm \sqrt{49 + 32}}{8}
\]

\[
y = \frac{-7 \pm \sqrt{81}}{8}
\]

\[
y = \frac{-7 \pm 9}{8}
\]

This gives two solutions:
1. \(y = \frac{-7 + 9}{8} = \frac{2}{8} = \frac{1}{4}\)
2. \(y = \frac{-7 - 9}{8} = \frac{-16}{8} = -2\)

Since \(y = 2^x\), and \(2^x\) cannot be negative, we discard \(y = -2\).

### Step 4: Solve for \(x\).
Now, we have \(y = \frac{1}{4}\), which means:
\[
2^x = \frac{1}{4}
\]

We can rewrite \(\frac{1}{4}\) as \(2^{-2}\), so:
\[
2^x = 2^{-2}
\]

Therefore:
\[
x = -2
\]

### Final Answer
\[
x = -2
\]

Would you like more details on any step or have further questions?

---

**Related Questions:**

1. How do we solve similar equations involving exponents and logarithms?
2. How can we verify the solution by plugging \(x = -2\) back into the original equation?
3. What are the rules for manipulating exponents and powers?
4. How would this equation change if it had a different base, like \(3^x\)?
5. Why can \(2^x\) never be negative?

**Tip:** In exponential equations, substituting expressions with a single variable, like \(y = 2^x\), can often simplify the problem into a manageable form.

Learn how to solve the exponential equation (4^x+1)=2-7*2^x by rewriting terms and applying substitution and the quadratic formula. This step-by-step guide explores key exponential and algebraic concepts, making it suitable for students in Grades 9-11.

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