Certainly! Let's solve the equation [tex]\(\log_3(x+1) + \log_3(x-1) = 2\)[/tex] step by step.
1. Combine the Logarithms:
Using the property of logarithms that [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex], we can combine the left side of the equation:
[tex]\[
\log_3((x+1)(x-1)) = 2
\][/tex]
2. Simplify the Argument:
Next, simplify the argument of the logarithm by multiplying the terms:
[tex]\[
(x+1)(x-1) = x^2 - 1
\][/tex]
So our equation now becomes:
[tex]\[
\log_3(x^2 - 1) = 2
\][/tex]
3. Rewrite in Exponential Form:
Express the logarithm equation as an exponential equation. Recall that if [tex]\(\log_b(a) = c\)[/tex], then [tex]\(a = b^c\)[/tex]. Thus:
[tex]\[
x^2 - 1 = 3^2
\][/tex]
Simplify [tex]\(3^2\)[/tex]:
[tex]\[
x^2 - 1 = 9
\][/tex]
4. Solve for [tex]\(x\)[/tex]:
To find [tex]\(x\)[/tex], add 1 to both sides:
[tex]\[
x^2 = 10
\][/tex]
Now, take the square root of both sides:
[tex]\[
x = \pm \sqrt{10}
\][/tex]
5. Verify Solutions:
It's important to check whether both solutions are valid in the context of the original logarithmic equation because the arguments of the logarithms must be positive.
- For [tex]\(x = \sqrt{10}\)[/tex]:
[tex]\[
x + 1 = \sqrt{10} + 1 > 0 \quad \text{and} \quad x - 1 = \sqrt{10} - 1 > 0
\][/tex]
Since both values are positive, [tex]\(x = \sqrt{10}\)[/tex] is valid.
- For [tex]\(x = -\sqrt{10}\)[/tex]:
[tex]\[
x + 1 = -\sqrt{10} + 1 < 0 \quad \text{(invalid because argument of log must be positive)}
\][/tex]
Thus, [tex]\(x = -\sqrt{10}\)[/tex] is not valid because it causes the arguments of the logarithms to be non-positive.
Therefore, the only valid solution to the equation is:
[tex]\[
x = \sqrt{10}
\][/tex]
