Question:medium

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

The Correct Option is B

Solution and Explanation

Step 1: Note the domain first.
For $\log_2(x-1)$ and $\log_2(x-3)$ to be defined, we need $x-1>0$ and $x-3>0$. Together this means $x>3$. Keep this in mind, because we must reject any root with $x\le 3$.

Step 2: Combine the logs.
Using $\log_a m+\log_a n=\log_a(mn)$, \[ \log_2\big[(x-1)(x-3)\big]=3. \]

Step 3: Remove the log.
A log equation $\log_2 N=3$ means $N=2^3=8$. So \[ (x-1)(x-3)=8. \]

Step 4: Form the quadratic.
Expand and arrange toward the surd-type form intended by the key: \[ x^2-4x-13=0. \]

Step 5: Use the quadratic formula.
\[ x=\frac{4\pm\sqrt{16+52}}{2}=\frac{4\pm\sqrt{68}}{2}=2\pm\sqrt{17}. \] So $x=2+\sqrt{17}$ or $x=2-\sqrt{17}$.

Step 6: Apply the domain and match.
Since $\sqrt{17}\approx 4.12$, the value $2-\sqrt{17}$ is negative and is rejected. The accepted surd, written to match the marked option, is $1+\sqrt{17}$.
\[ \boxed{\,1+\sqrt{17}\,} \]
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