We are given the equation:

\[
2 \cdot 4^{(x+1)} + 2^{(x+2)} = \frac{2}{3}
\]

### Step 1: Express \(4^{(x+1)}\) in terms of base 2.
Since \(4 = 2^2\), we can rewrite \(4^{(x+1)}\) as:

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

So, the original equation becomes:

\[
2 \cdot 2^{2x+2} + 2^{x+2} = \frac{2}{3}
\]

### Step 2: Simplify the equation.
Using the properties of exponents, simplify each term:

\[
2 \cdot 2^{2x+2} = 2^{2x+3}
\]
\[
2^{x+2} = 2^{x+2}
\]

Thus, the equation becomes:

\[
2^{2x+3} + 2^{x+2} = \frac{2}{3}
\]

### Step 3: Introduce substitution for simplification.
Let \( y = 2^x \), then the equation becomes:

\[
2^{2x+3} = 8y^2 \quad \text{and} \quad 2^{x+2} = 4y
\]

Now, the equation is:

\[
8y^2 + 4y = \frac{2}{3}
\]

### Step 4: Eliminate the fraction.
Multiply the entire equation by 3 to eliminate the fraction:

\[
3(8y^2 + 4y) = 2
\]

\[
24y^2 + 12y = 2
\]

### Step 5: Set up a quadratic equation.
Move all terms to one side of the equation:

\[
24y^2 + 12y - 2 = 0
\]

### Step 6: Solve the quadratic equation.
Use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 24\), \(b = 12\), and \(c = -2\). Plugging these into the formula:

\[
y = \frac{-12 \pm \sqrt{12^2 - 4(24)(-2)}}{2(24)}
\]
\[
y = \frac{-12 \pm \sqrt{144 + 192}}{48}
\]
\[
y = \frac{-12 \pm \sqrt{336}}{48}
\]
\[
y = \frac{-12 \pm 18.33}{48}
\]

This gives two solutions:

\[
y_1 = \frac{-12 + 18.33}{48} \approx \frac{6.33}{48} \approx 0.132
\]
\[
y_2 = \frac{-12 - 18.33}{48} \approx \frac{-30.33}{48} \approx -0.632
\]

### Step 7: Back-substitute for \(y = 2^x\).
Since \(y = 2^x\), we solve for \(x\):

For \(y_1 \approx 0.132\):

\[
2^x = 0.132 \implies x \approx \log_2(0.132) \approx -2.92
\]

For \(y_2 \approx -0.632\), there is no solution since \(2^x\) cannot be negative.

### Final Answer:
The solution is:

\[
x \approx -2.92
\]

Would you like further details or have any questions? Here are some related questions to explore:

1. What are the properties of logarithms used in solving exponential equations?
2. How do we derive the quadratic formula?
3. What are the conditions for a quadratic equation to have real solutions?
4. How do transformations of exponential functions work graphically?
5. How can logarithms be used to solve other types of exponential equations?

**Tip:** Always check if your solution makes sense in the context of the equation, especially when dealing with negative results for exponential functions.

Step-by-step solution to solve the exponential equation 2 * 4^(x+1) + 2^(x+2) = 2/3 using substitution and the quadratic formula. Learn how to express exponents in base 2 and use logarithms to find the solution, suitable for students in Grades 10-12.

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