Question

If \( \log_{10}(x+1) + \log_{10}(x-1) = 1 \), then \( x = \)

• A. 3
• B. \(\sqrt{10}\)
• C. \(\sqrt{11}\)
• D. 10

Hint:

Check domain after solving logarithmic equations.

Answer 1

The Correct Option is B

To solve the given equation \( \log_{10}(x+1) + \log_{10}(x-1) = 1 \), we use the logarithmic property: \[ \log_b(m) + \log_b(n) = \log_b(mn) \]

Thus,

\[ \log_{10}((x+1)(x-1)) = 1 \]

Simplifying inside the logarithm:

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

So the equation becomes:

\[ \log_{10}(x^2 - 1) = 1 \]

Using the definition of logarithms, if \( \log_{10}(y) = 1 \), then \( y = 10 \). Hence:

\[ x^2 - 1 = 10 \]

Solving for \(x\):

\[ x^2 = 11 \] \[ x = \pm \sqrt{11} \].

Since both \( \log_{10}(x+1) \) and \( \log_{10}(x-1) \) are defined only when: \[ x+1 > 0 \quad \text{and} \quad x-1 > 0 \] which gives \(x > 1\), we discard the negative solution.

Therefore, the valid solution is: \[ x = \sqrt{11} \]

Correct Answer: \( \sqrt{11} \)