The equation in the image is:

$\log_3 (x - 1) + \log_3 (x + 1) = 2$

Step-by-Step Solution:

1. Use the logarithm addition property: The sum of two logarithms with the same base can be rewritten as a single logarithm. This property is: $\log_b A + \log_b B = \log_b (A \times B)$ Applying this to our equation: $\log_3 \left((x - 1)(x + 1)\right) = 2$

2. Simplify the expression inside the logarithm: $(x - 1)(x + 1)$ is a difference of squares. Using the identity $a^2 - b^2 = (a - b)(a + b)$, we get: $(x - 1)(x + 1) = x^2 - 1$ Now the equation becomes: $\log_3 (x^2 - 1) = 2$

3. Eliminate the logarithm: To remove the logarithm, rewrite the equation in exponential form. The logarithmic equation $\log_b A = C$ is equivalent to the exponential equation $A = b^C$. Applying this to our equation: $x^2 - 1 = 3^2$ $x^2 - 1 = 9$

4. Solve for $x$: Add 1 to both sides: $x^2 = 9 + 1 = 10$ Take the square root of both sides: $x = \pm \sqrt{10}$ So, the possible solutions are: $x = \sqrt{10} \quad \text{or} \quad x = -\sqrt{10}$

5. Check for domain restrictions: Since we have logarithmic terms $\log_3 (x - 1)$ and $\log_3 (x + 1)$, the arguments inside the logarithms must be positive. Therefore: $x - 1 > 0 \quad \Rightarrow \quad x > 1$ $x + 1 > 0 \quad \Rightarrow \quad x > -1$ The stricter condition is $x > 1$, meaning only $x = \sqrt{10}$ is valid, while $x = -\sqrt{10}$ is not valid because it does not satisfy $x > 1$.

Final Answer:

$x = \sqrt{10}$

Let me know if you need further clarification or have any specific questions! Here are some follow-up questions to expand your understanding:

1. Can you rewrite other logarithmic sums using the product rule?
2. What happens if the base of the logarithms is not the same?
3. Can you explain how domain restrictions affect logarithmic equations?
4. What if the right side of the equation was a different number, say 3, instead of 2?
5. How would you solve this if it had subtraction instead of addition between the logs?

Tip:

Always check the domain of logarithmic expressions before accepting solutions—logarithms are only defined for positive arguments!