https://file.aiquickdraw.com/mathbot/file/1727420016114h72n7cgs.jpg

The problem in the image is solving the equation \( 3^{2x} - 12 \cdot 3^x + 27 = 0 \).

Let's work through the steps:

### Step 1: Let \( a = 3^x \)
This substitution simplifies the equation. We know that \( 3^{2x} = (3^x)^2 \), so the equation becomes:
\[
a^2 - 12a + 27 = 0
\]
This is a quadratic equation in terms of \( a \).

### Step 2: Solve the quadratic equation
To solve \( a^2 - 12a + 27 = 0 \), we can use the quadratic formula:
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -12 \), and \( c = 27 \). Plugging these into the formula:
\[
a = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(27)}}{2(1)}
\]
\[
a = \frac{12 \pm \sqrt{144 - 108}}{2}
\]
\[
a = \frac{12 \pm \sqrt{36}}{2}
\]
\[
a = \frac{12 \pm 6}{2}
\]
This gives two solutions:
\[
a = \frac{12 + 6}{2} = 9 \quad \text{or} \quad a = \frac{12 - 6}{2} = 3
\]
So, \( a = 9 \) or \( a = 3 \).

### Step 3: Solve for \( x \)
Recall that \( a = 3^x \), so we now solve:
1. \( 3^x = 9 \). Since \( 9 = 3^2 \), we have \( x = 2 \).
2. \( 3^x = 3 \). Since \( 3 = 3^1 \), we have \( x = 1 \).

### Conclusion:
The solutions to the equation \( 3^{2x} - 12 \cdot 3^x + 27 = 0 \) are \( x = 2 \) and \( x = 1 \).

---

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

Here are 5 related questions for practice:
1. Solve the equation \( 4^{2x} - 8 \cdot 4^x + 16 = 0 \).
2. How do you solve a quadratic equation using factoring?
3. Solve \( 5^{2x} - 10 \cdot 5^x + 25 = 0 \).
4. What is the quadratic formula and when is it used?
5. Solve \( 2^{2x} - 3 \cdot 2^x - 4 = 0 \).

**Tip:** When dealing with exponential equations, substituting \( a = b^x \) can simplify them into quadratic equations.

Learn how to solve the exponential equation 3^(2x) - 12 * 3^x + 27 = 0 using substitution and the quadratic formula. Detailed step-by-step solution, covering key concepts like exponential and quadratic equations, suitable for Grades 9-11.

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