Log (x - 1) + log (x + 1) = 1, find the value of x.

We are given the equation:

\[
\log(x - 1) + \log(x + 1) = 1
\]

### Step 1: Use the logarithmic property
We can combine the logarithms using the property of logarithms:

\[
\log(a) + \log(b) = \log(a \cdot b)
\]

So, applying this to the given equation:

\[
\log((x - 1)(x + 1)) = 1
\]

### Step 2: Simplify the expression
The product \((x - 1)(x + 1)\) simplifies to:

\[
(x - 1)(x + 1) = x^2 - 1
\]

Now, we have:

\[
\log(x^2 - 1) = 1
\]

### Step 3: Convert from logarithmic form to exponential form
Recall that \(\log_b(y) = z\) implies \(y = b^z\). Here, since we are using the common logarithm (base 10), we convert the equation to:

\[
x^2 - 1 = 10^1
\]

\[
x^2 - 1 = 10
\]

### Step 4: Solve for \(x\)
Now, solve the equation:

\[
x^2 = 10 + 1 = 11
\]

\[
x = \pm \sqrt{11}
\]

### Step 5: Check for validity
Since the logarithmic function is defined only for positive arguments, we must ensure that both \(x - 1 > 0\) and \(x + 1 > 0\). This implies:

\[
x - 1 > 0 \quad \text{or} \quad x > 1
\]

Therefore, \(x = \sqrt{11}\) is the only valid solution (as \(x = -\sqrt{11}\) would make \(\log(x - 1)\) undefined).

### Final Answer:
The value of \(x\) is:

\[
x = \sqrt{11}
\]

Would you like more details on any step?

Here are 5 related questions to practice:
1. Solve \(\log(x + 3) + \log(x - 2) = 2\).
2. Solve \(\log(2x) + \log(3) = 1\).
3. If \(\log(x^2 - 4) = 0\), find \(x\).
4. Solve \(\log(x + 5) - \log(x - 1) = 1\).
5. Find \(x\) if \(\log(x^2 - 9) = 1\).

**Tip:** Always check the domain restrictions when solving logarithmic equations, as logarithms are only defined for positive values.

Learn how to solve the logarithmic equation Log(x - 1) + Log(x + 1) = 1 step-by-step. This problem covers logarithmic properties and exponential equations, suitable for high school students in Grades 9-12.

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