Solve 9$$ ^{x} $$– 12(3$$ ^{x} $$) + 27 = 0 for x

To solve the equation $9^x - 12 \cdot 3^x + 27 = 0$ for $x$, we can start by expressing $9^x$ in terms of $3^x$. 

Recall that $9 = 3^2$. Therefore, $9^x = (3^2)^x = 3^{2x}$.

Let $y = 3^x$. Then $9^x = 3^{2x} = y^2$.

So, the original equation becomes:
\[ y^2 - 12y + 27 = 0 \]

This is a quadratic equation in $y$. To solve for $y$, we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, $a = 1$, $b = -12$, and $c = 27$. Plugging these values into the formula gives:
\[ y = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} \]
\[ y = \frac{12 \pm \sqrt{144 - 108}}{2} \]
\[ y = \frac{12 \pm \sqrt{36}}{2} \]
\[ y = \frac{12 \pm 6}{2} \]

This gives us two solutions for $y$:
\[ y = \frac{12 + 6}{2} = 9 \]
\[ y = \frac{12 - 6}{2} = 3 \]

Recall that $y = 3^x$. Therefore, we have:
\[ 3^x = 9 \]
\[ 3^x = 3^2 \]
\[ x = 2 \]

And:
\[ 3^x = 3 \]
\[ 3^x = 3^1 \]
\[ x = 1 \]

So, the solutions to the equation $9^x - 12 \cdot 3^x + 27 = 0$ are:
\[ x = 1 \text{ and } x = 2 \]

Would you like more details or have any questions?

Here are 5 questions you might be interested in:
1. How do you solve quadratic equations?
2. What is the quadratic formula and how is it derived?
3. How do you solve exponential equations?
4. Can you explain the properties of exponents?
5. What are some common methods for solving algebraic equations?

**Tip:** When solving equations involving exponents, it's often helpful to rewrite the terms with a common base to simplify the equation.

Learn how to solve the equation 9^x - 12 * 3^x + 27 = 0 step-by-step with detailed explanations and solutions. This problem involves exponential and quadratic equations, requiring understanding of the quadratic formula. Suitable for high school students in Grades 10-12.

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