\log _4\left(x-3\right)+\log _4\left(x+3\right)=2

We are given the equation:

\[
\log_4(x - 3) + \log_4(x + 3) = 2
\]

To solve for \(x\), follow these steps:

### Step 1: Apply the logarithmic property
Use the property of logarithms that states:

\[
\log_b(A) + \log_b(B) = \log_b(A \cdot B)
\]

This allows us to combine the two logarithms on the left-hand side:

\[
\log_4\left((x - 3)(x + 3)\right) = 2
\]

### Step 2: Simplify the expression inside the logarithm
Notice that \((x - 3)(x + 3)\) is a difference of squares:

\[
(x - 3)(x + 3) = x^2 - 9
\]

Thus, the equation becomes:

\[
\log_4(x^2 - 9) = 2
\]

### Step 3: Rewrite the equation in exponential form
Recall that \(\log_b(A) = C\) is equivalent to \(b^C = A\). Applying this property, we convert the logarithmic equation to an exponential

\log_4\left(x - 3\right) + \log_4\left(x + 3\right) = 2

Learn how to solve the logarithmic equation \log_4(x - 3) + \log_4(x + 3) = 2 using logarithmic properties and exponential equations. This problem explores key concepts such as the difference of squares and logarithmic properties, suitable for students in Grades 9-12.

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