Question

If \[ \log_2(x-1)+\log_2(x+1)=3, \] then \(x\) is equal to:

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

After solving any logarithmic equation, always check whether the obtained solutions make every logarithmic argument positive. This step prevents acceptance of extraneous roots.

Answer 1

The Correct Option is B

Concept: Logarithmic equations are usually solved by first combining logarithms using logarithmic identities. Once combined, the logarithmic equation can be transformed into an equivalent exponential equation. A very important aspect of solving logarithmic equations is checking the domain restrictions. Any value that makes the logarithmic argument zero or negative must be rejected, even if it satisfies the algebraic equation obtained later.

Step 1: Combine the logarithms. Using \[ \log_a m+\log_a n = \log_a(mn), \] we obtain \[ \log_2[(x-1)(x+1)] = 3. \]

Step 2: Simplify the expression inside the logarithm. Using the identity \[ (a-b)(a+b)=a^2-b^2, \] we get \[ \log_2(x^2-1)=3. \]

Step 3: Convert into exponential form. Since \[ \log_2(x^2-1)=3, \] we have \[ x^2-1=2^3. \] \[ x^2-1=8. \] \[ x^2=9. \]

Step 4: Solve the quadratic equation. Taking square roots, \[ x=\pm3. \] Thus the algebraic solutions are \[ x=3 \] and \[ x=-3. \]

Step 5: Apply the domain restrictions. The logarithmic arguments must be positive. Therefore, \[ x-1>0 \] which gives \[ x>1. \] Hence \[ x=-3 \] is not admissible. Only \[ x=3 \] satisfies the domain condition.

Step 6: Final Conclusion. Therefore, \[ \boxed{x=3} \] Hence the correct answer is \[ \boxed{\text{Option (B)}}. \]