We are given the equation:

\[
2 \log_3{x} - \log_3{(x+4)} = 2
\]

### Step 1: Apply the properties of logarithms
We can use the **power rule** of logarithms to simplify the first term:

\[
2 \log_3{x} = \log_3{x^2}
\]

Thus, the equation becomes:

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

### Step 2: Apply the quotient rule of logarithms
The **quotient rule** of logarithms states that:

\[
\log_b{A} - \log_b{B} = \log_b{\left(\frac{A}{B}\right)}
\]

Applying this to the equation, we get:

\[
\log_3\left(\frac{x^2}{x+4}\right) = 2
\]

### Step 3: Convert the logarithmic equation to an exponential form
Using the definition of logarithms, \(\log_b{A} = C\) implies \(A = b^C\), we convert the equation to:

\[
\frac{x^2}{x+4} = 3^2
\]

This simplifies to:

\[
\frac{x^2}{x+4} = 9
\]

### Step 4: Solve the resulting quadratic equation
Multiply both sides by \(x + 4\) to eliminate the fraction:

\[
x^2 = 9(x + 4)
\]

Expand the right-hand side:

\[
x^2 = 9x + 36
\]

Now, bring all terms to one side:

\[
x^2 - 9x - 36 = 0
\]

### Step 5: Solve the quadratic equation
We can solve this using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -9\), and \(c = -36\).

\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(-36)}}{2(1)}
\]
\[
x = \frac{9 \pm \sqrt{81 + 144}}{2}
\]
\[
x = \frac{9 \pm \sqrt{225}}{2}
\]
\[
x = \frac{9 \pm 15}{2}
\]

Thus, we have two solutions:

\[
x = \frac{9 + 15}{2} = 12 \quad \text{or} \quad x = \frac{9 - 15}{2} = -3
\]

### Step 6: Check for extraneous solutions
Logarithms are only defined for positive arguments, so \(x + 4 > 0\) implies \(x > -4\). However, \(x = -3\) would make \(\log_3{(-3)}\) undefined, so we discard \(x = -3\).

Thus, the only valid solution is:

\[
\boxed{x = 12}
\]

Do you have any questions or would you like more details?

Here are 5 related questions:
1. How do you solve logarithmic equations using the product and quotient rules?
2. What happens when a logarithmic equation has no real solution?
3. How do you determine the domain restrictions in logarithmic functions?
4. Can you explain how to solve a system involving exponential and logarithmic equations?
5. How does the quadratic formula work, and when is it used?

**Tip:** Always check if your solution is valid by substituting back into the original logarithmic equation, especially when dealing with logarithms.

This step-by-step guide explains how to solve the equation 2log3x – log3(x + 4) = 2. The solution involves applying the properties of logarithms and solving the resulting quadratic equation, suitable for Grades 10-12 students.

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