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We are given the equation:

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

To solve for \( x \), let's proceed step by step.

### Step 1: Convert the logarithmic equation into exponential form

The equation \(\log_b(a) = c\) is equivalent to \(b^c = a\).

In this case, we have:

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

This converts to:

\[
3^1 = 4x - x^2
\]

### Step 2: Simplify the equation

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

Rearrange this into a standard quadratic form:

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

### Step 3: Solve the quadratic equation

This is a simple quadratic equation, and we can factor it:

\[
(x - 1)(x - 3) = 0
\]

So, the solutions are:

\[
x = 1 \quad \text{or} \quad x = 3
\]

### Step 4: Check for extraneous solutions

Since the original equation involves a logarithm, we need to ensure that the argument of the logarithm is positive. That is, \(4x - x^2 > 0\).

For \(x = 1\):

\[
4(1) - 1^2 = 4 - 1 = 3 > 0
\]

For \(x = 3\):

\[
4(3) - 3^2 = 12 - 9 = 3 > 0
\]

Both solutions are valid.

### Final Answer:

The solutions are \(x = 1\) and \(x = 3\).

Would you like a more detailed explanation of any step?

Here are 5 related questions for further exploration:
1. How do you solve logarithmic equations with different bases?
2. Can logarithms have negative arguments? Why or why not?
3. How can you solve a quadratic equation using the quadratic formula?
4. What is the general rule for converting logarithmic equations to exponential form?
5. What are the properties of logarithms that are useful in solving equations?

**Tip**: When solving logarithmic equations, always check the domain of the original logarithmic expression to ensure your solutions are valid!

Learn how to solve the logarithmic equation log3(4x - x^2) = 1, converting it into a quadratic form and solving using factoring. This problem demonstrates key concepts in logarithms and quadratic equations, suitable for Grades 10-12.

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