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Solve the logarithmic equation. Express all solutions.

[tex]\[ \log_3\left(x^2 - 9\right) - \log_3(x + 3) = 1 \][/tex]

Select the correct choice below and, if necessary, fill in the answer box to complete your choice:

A. [tex]\( x = \boxed{} \)[/tex]

B. There is no solution.

Sagot :

GinnyAnswer
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]
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