Solve the equation 𝑙𝑜𝑔3(𝑥^2 − 4𝑥 + 4) = 1 + 𝑙𝑜𝑔3(𝑥 − 2).

We are given the equation:

\[
\log_3(x^2 - 4x + 4) = 1 + \log_3(x - 2)
\]

### Step 1: Simplify the equation
We can rewrite the equation using the logarithmic properties.

The right-hand side is:

\[
1 + \log_3(x - 2)
\]

We can express 1 as \(\log_3(3)\), so the right-hand side becomes:

\[
\log_3(3) + \log_3(x - 2)
\]

Using the logarithmic property \(\log_b(a) + \log_b(c) = \log_b(ac)\), the right-hand side simplifies to:

\[
\log_3(3(x - 2))
\]

Thus, the equation becomes:

\[
\log_3(x^2 - 4x + 4) = \log_3(3(x - 2))
\]

### Step 2: Remove the logarithms
Since the logarithms on both sides have the same base (base 3), we can remove the \(\log_3\) terms and set the arguments equal to each other:

\[
x^2 - 4x + 4 = 3(x - 2)
\]

### Step 3: Solve the quadratic equation
Now expand and simplify the equation:

\[
x^2 - 4x + 4 = 3x - 6
\]

Bring all terms to one side:

\[
x^2 - 4x + 4 - 3x + 6 = 0
\]

Simplify:

\[
x^2 - 7x + 10 = 0
\]

### Step 4: Factor the quadratic equation
We factor the quadratic equation:

\[
x^2 - 7x + 10 = (x - 5)(x - 2) = 0
\]

### Step 5: Solve for \(x\)
From the factored equation, the solutions are:

\[
x - 5 = 0 \quad \text{or} \quad x - 2 = 0
\]

Thus, \(x = 5\) or \(x = 2\).

### Step 6: Check for extraneous solutions
We need to check whether any of the solutions make the argument of the logarithm negative or zero, as logarithms are undefined for such values.

- For \(x = 5\):
  - \(\log_3(5^2 - 4 \cdot 5 + 4) = \log_3(9)\), which is defined.
  - \(\log_3(5 - 2) = \log_3(3)\), which is defined.

- For \(x = 2\):
  - \(\log_3(2^2 - 4 \cdot 2 + 4) = \log_3(0)\), which is undefined.

Thus, \(x = 2\) is an extraneous solution.

### Final Answer:
The only valid solution is:

\[
x = 5
\]

Would you like more details or clarifications on any part of the solution?

Here are 5 related questions to deepen your understanding:
1. What are the properties of logarithms that were used in this problem?
2. How can you solve logarithmic equations with different bases?
3. How do you determine if a solution is extraneous in logarithmic equations?
4. Can the quadratic formula be used to solve all quadratic equations?
5. How do logarithmic and exponential functions relate to each other?

**Tip:** Always check for extraneous solutions when solving logarithmic equations, as they can arise when the argument of the logarithm becomes zero or negative.

Solve the equation log₃(x² − 4x + 4) = 1 + log₃(x − 2).

Explore how to solve the logarithmic equation log₃(x² − 4x + 4) = 1 + log₃(x − 2), incorporating algebraic and logarithmic concepts. This step-by-step solution includes properties of logarithms and quadratic equations, suitable for students in Grades 9-11.

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