We are given the equation:

$\log_4(x - 3) + \log_4(x + 3) = 2$

Step 1: Combine the logarithmic terms

Using the logarithmic property:

$\log_b(a) + \log_b(c) = \log_b(a \cdot c)$

We combine the two logarithms:

$\log_4((x - 3)(x + 3)) = 2$

Step 2: Expand the product inside the logarithm

Using the difference of squares formula:

$(x - 3)(x + 3) = x^2 - 9$

Now the equation becomes:

$\log_4(x^2 - 9) = 2$

Step 3: Rewrite the equation in exponential form

The logarithmic equation $\log_b(y) = c$ can be rewritten as $b^c = y$. Applying this property:

$x^2 - 9 = 4^2$

Since $4^2 = 16$, we have:

$x^2 - 9 = 16$

Step 4: Solve for $x$

Now solve the equation:

$x^2 = 16 + 9 = 25$

$x = \pm \sqrt{25}$

$x = \pm 5$

Step 5: Check for extraneous solutions

Since logarithms are only defined for positive arguments, we must check both solutions in the original equation:

• For $x = 5$: $\log_4(5 - 3) + \log_4(5 + 3) = \log_4(2) + \log_4(8)$ This is valid since both arguments are positive.

• For $x = -5$: $\log_4(-5 - 3) = \log_4(-8)$ This is undefined because the logarithm of a negative number is not real.

Therefore, the only valid solution is:

$x = 5$

Final Answer:

$x = 5$

Would you like any more details or have any questions?

Here are 5 related questions you might find interesting:

1. How do you solve logarithmic equations with different bases?
2. What are the properties of logarithms used in solving equations?
3. How does the change of base formula work for logarithms?
4. What is the domain of logarithmic functions?
5. How do you solve exponential equations that involve logarithms?

Tip: Always check for extraneous solutions in logarithmic equations, since the arguments of logarithms must be positive.