Question:
If
\[
\log_2(x-1)+\log_2(x-3)=3,
\]
then the value of \(x\) is
If
\[
\log_2(x-1)+\log_2(x-3)=3,
\]
then the value of \(x\) is
Show Hint
Always check the domain after solving logarithmic equations. Many extraneous roots arise because logarithms are defined only for positive arguments.
Updated On: Jun 10, 2026
- \(5\)
- \(1+\sqrt{17}\)
- \(4+\sqrt{5}\)
- \(3+\sqrt{17}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The logarithmic identity
\[
\log_a m+\log_a n=\log_a(mn)
\]
allows us to combine logarithms having the same base.
Step 1: Combine the logarithms \[ \log_2[(x-1)(x-3)]=3. \]
Step 2: Convert to exponential form \[ (x-1)(x-3)=2^3. \] \[ (x-1)(x-3)=8. \]
Step 3: Expand \[ x^2-4x+3=8. \] \[ x^2-4x-5=0. \]
Step 4: Solve the quadratic \[ x=\frac{4\pm\sqrt{16+20}}{2}. \] \[ x=\frac{4\pm6}{2}. \] Thus, \[ x=5 \] or \[ x=-1. \]
Step 5: Apply the domain restriction Since \[ x-3>0, \] we must have \[ x>3. \] Therefore, \[ x=5. \] Hence, \[ \boxed{5}. \]
Step 1: Combine the logarithms \[ \log_2[(x-1)(x-3)]=3. \]
Step 2: Convert to exponential form \[ (x-1)(x-3)=2^3. \] \[ (x-1)(x-3)=8. \]
Step 3: Expand \[ x^2-4x+3=8. \] \[ x^2-4x-5=0. \]
Step 4: Solve the quadratic \[ x=\frac{4\pm\sqrt{16+20}}{2}. \] \[ x=\frac{4\pm6}{2}. \] Thus, \[ x=5 \] or \[ x=-1. \]
Step 5: Apply the domain restriction Since \[ x-3>0, \] we must have \[ x>3. \] Therefore, \[ x=5. \] Hence, \[ \boxed{5}. \]
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