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We are given the equation:

\[
\log_5(x^2) = \log_5(3x - 2)
\]

### Step 1: Remove the logarithms

Since the base of the logarithm on both sides is the same (base 5), we can equate the arguments inside the logarithms:

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

### Step 2: Solve the quadratic equation

Rearrange the equation into standard quadratic form:

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

Now, factor the quadratic:

\[
(x - 1)(x - 2) = 0
\]

### Step 3: Solve for \(x\)

Set each factor equal to zero:

\[
x - 1 = 0 \quad \text{or} \quad x - 2 = 0
\]

\[
x = 1 \quad \text{or} \quad x = 2
\]

### Step 4: Check for extraneous solutions

For logarithmic equations, it's important to check that the solutions do not result in taking the logarithm of a negative number or zero.

- For \(x = 1\): \(3(1) - 2 = 1\), so both sides of the original equation are valid.
- For \(x = 2\): \(3(2) - 2 = 4\), so both sides of the original equation are valid.

Thus, both solutions are valid.

### Final Answer:

The solutions are \(x = 1\) and \(x = 2\).

---

Would you like further clarification on this? Here are some related questions:

1. How do you solve quadratic equations by factoring?
2. What are extraneous solutions in logarithmic equations?
3. How do logarithmic properties help in solving equations?
4. Can logarithms be negative? Why or why not?
5. How does the base of a logarithm affect solving equations?

**Tip:** Always remember to check for extraneous solutions when dealing with logarithmic equations!

Learn how to solve the logarithmic equation log_5(x^2) = log_5(3x - 2) by removing the logarithms and solving the resulting quadratic equation. This problem covers key algebraic concepts, including solving logarithmic and quadratic equations. Suitable for students in Grades 9-12.

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