To solve the logarithmic equation [tex]\(\log_3(x^2 - 9) - \log_3(x + 3) = 1\)[/tex], let's follow a step-by-step approach using properties of logarithms.
### Step 1: Apply Logarithm Properties
First, we can use the property of logarithms that states [tex]\(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\)[/tex]. Applying this property to our equation, we get:
[tex]\[
\log_3\left(\frac{x^2 - 9}{x + 3}\right) = 1
\][/tex]
### Step 2: Exponential Form
Next, we convert the logarithmic equation to its exponential form. Recall that [tex]\(\log_b(A) = C\)[/tex] is equivalent to [tex]\(A = b^C\)[/tex]. Therefore:
[tex]\[
\frac{x^2 - 9}{x + 3} = 3^1
\][/tex]
This simplifies to:
[tex]\[
\frac{x^2 - 9}{x + 3} = 3
\][/tex]
### Step 3: Solve the Rational Equation
To eliminate the denominator, we multiply both sides of the equation by [tex]\(x + 3\)[/tex]:
[tex]\[
x^2 - 9 = 3(x + 3)
\][/tex]
### Step 4: Simplify the Equation
Expand and simplify the equation:
[tex]\[
x^2 - 9 = 3x + 9
\][/tex]
[tex]\[
x^2 - 3x - 18 = 0
\][/tex]
### Step 5: Solve the Quadratic Equation
Now, we solve the quadratic equation [tex]\(x^2 - 3x - 18 = 0\)[/tex]. We can do this by factoring:
[tex]\[
(x - 6)(x + 3) = 0
\][/tex]
### Step 6: Find the Critical Points
Set each factor equal to zero to find the potential solutions:
[tex]\[
x - 6 = 0 \quad \text{or} \quad x + 3 = 0
\][/tex]
[tex]\[
x = 6 \quad \text{or} \quad x = -3
\][/tex]
### Step 7: Check for Validity
Finally, we need to check if these solutions are valid within the context of the original logarithmic equation. Logarithms are defined only for positive arguments, so we must ensure that both [tex]\(x^2 - 9 > 0\)[/tex] and [tex]\(x + 3 > 0\)[/tex].
1. For [tex]\(x = 6\)[/tex]:
[tex]\[
x^2 - 9 = 36 - 9 = 27 > 0 \quad \text{and} \quad x + 3 = 6 + 3 = 9 > 0
\][/tex]
So, [tex]\(x = 6\)[/tex] is valid.
2. For [tex]\(x = -3\)[/tex]:
[tex]\[
x^2 - 9 = (-3)^2 - 9 = 9 - 9 = 0 \quad \text{and} \quad x + 3 = -3 + 3 = 0
\][/tex]
In both expressions, the arguments of the logarithms become zero, which is not allowed. Therefore, [tex]\(x = -3\)[/tex] is not a valid solution.
### Conclusion
The only valid solution to the equation is:
[tex]\[
\boxed{6}
\][/tex]
