https://file.aiquickdraw.com/mathbot/file/1725849089059i37b0iek.jpg

Let's analyze and solve the two given equations from the image:

### 1. Problem 2:
\[
2 \cdot 4^x - 5 \cdot 2^x + 2 - 6 = 0
\]

### 2. Problem 3:
\[
9^x - 2 \cdot 3^{x+1} - 27 = 0
\]

I'll start by solving the first one and then the second.

---

### Solution to Problem 2:
We can rewrite \(4^x\) as \((2^2)^x = 2^{2x}\), and solve the equation accordingly.

\[
2 \cdot 2^{2x} - 5 \cdot 2^x + 2 - 6 = 0
\]
Simplifying the constant terms:
\[
2 \cdot 2^{2x} - 5 \cdot 2^x - 4 = 0
\]
Let \( y = 2^x \), so the equation becomes:
\[
2y^2 - 5y - 4 = 0
\]
Now, we can solve this quadratic equation using the quadratic formula.

---

### Solution to Problem 3:
The second equation can be simplified using the fact that \(9^x = (3^2)^x = 3^{2x}\).

\[
3^{2x} - 2 \cdot 3^{x+1} - 27 = 0
\]
Let \( z = 3^x \), so the equation becomes:
\[
z^2 - 2 \cdot 3z - 27 = 0
\]
This is a quadratic equation in terms of \(z\), and we can solve it using the quadratic formula.

Let me calculate both quadratic equations and find the solutions.### Solutions:

#### 1. For Problem 2:
The quadratic equation \(2y^2 - 5y - 4 = 0\) has the following solutions:
\[
y_1 = \frac{5 - \sqrt{57}}{4}, \quad y_2 = \frac{5 + \sqrt{57}}{4}
\]
Since \(y = 2^x\), we'll need to check which of these solutions can be used to find a real value of \(x\), as \(2^x\) must be positive.

#### 2. For Problem 3:
The quadratic equation \(z^2 - 6z - 27 = 0\) gives:
\[
z_1 = -3, \quad z_2 = 9
\]
Since \(z = 3^x\), we can only consider the positive solution \(z = 9\), as \(3^x\) must also be positive. Therefore:
\[
3^x = 9 \quad \Rightarrow \quad x = 2
\]

Would you like me to proceed with simplifying the solutions for \(x\) in Problem 2, or do you have any other questions?

Here are some related questions for deeper understanding:
1. How does changing the base of exponential terms affect the solutions?
2. Why do we reject negative values for \(2^x\) and \(3^x\)?
3. What methods can be used to solve quadratic equations besides the quadratic formula?
4. How can logarithms be applied to find \(x\) after solving for \(y\)?
5. Can these types of exponential equations always be simplified using substitution?

Tip: When solving equations involving powers, always check if the base allows for negative values before discarding solutions.

Solve the equations: 1. 2 * 4^x - 5 * 2^x + 2 - 6 = 0  2. 9^x - 2 * 3^(x+1) - 27 = 0

Learn how to solve exponential equations like 2 * 4^x - 5 * 2^x + 2 - 6 = 0 and 9^x - 2 * 3^(x+1) - 27 = 0 by using substitution methods and applying the quadratic formula. Suitable for high school students.

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