Question

If $\log_2 (x-1) + \log_2 (x-3) = 3$, find the value(s) of $ x $.

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

Hint:

Tip: Always check domain restrictions after solving logarithmic equations.

Answer 1

The Correct Option is A

To solve the equation \(\log_2 (x-1) + \log_2 (x-3) = 3\), we start by using the property of logarithms that states \(\log_b a + \log_b c = \log_b (a \cdot c)\). Thus, we can rewrite the given equation as: 

\(\log_2 ((x-1)(x-3)) = 3\)

This can be further simplified by realizing that \(\log_2 y = 3\) implies \(y = 2^3\). Therefore:

\((x-1)(x-3) = 8\)

Next, expand the left-hand side:

\((x-1)(x-3) = x^2 - 3x - x + 3 = x^2 - 4x + 3\)

Equating both sides, we have:

\(x^2 - 4x + 3 = 8\)

Subtract 8 from both sides to set the quadratic equation to zero:

\(x^2 - 4x + 3 - 8 = 0\)

\(x^2 - 4x - 5 = 0\)

Now, solve this quadratic equation by factorization. We need two numbers whose product is \(-5\) and sum is \(-4\). These numbers are \(-5\) and \(1\). Thus, we can factor the quadratic as:

\((x-5)(x+1) = 0\)

Setting each factor equal to zero gives the solutions:

\(x-5=0\) or \(x+1=0\)

which implies:

\(x=5\) or \(x=-1\)

However, since logarithms are only defined for positive arguments, we must ensure that the values of \(x\) satisfy \(x-1>0\) and \(x-3>0\) which means \(x > 3\). Therefore, \(x = -1\) is not valid. So, the only valid solution is \(x = 5\).

Answer 2

Step 1: Use logarithm property 
\[ \log_2 (x-1) + \log_2 (x-3) = \log_2 \big[(x-1)(x-3)\big] = 3 \]

Step 2: Convert logarithmic equation to exponential form 
\[ (x-1)(x-3) = 2^3 = 8 \]

Step 3: Expand and form quadratic equation 
\[ x^2 - 4x + 3 = 8 \implies x^2 - 4x + 3 - 8 = 0 \implies x^2 - 4x - 5 = 0 \]

Step 4: Solve quadratic 
\[ x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2} \] \[ \Rightarrow x = \frac{4 + 6}{2} = 5 \quad \text{or} \quad x = \frac{4 - 6}{2} = -1 \]

Step 5: Check domain restrictions 
Since \(\log_2 (x-1)\) and \(\log_2 (x-3)\) are defined only if their arguments are positive: \[ x-1>0 \implies x>1, \quad x-3>0 \implies x>3 \] So \( x>3 \) must hold.

Step 6: Valid solution 
\( x = 5 \) is valid, \( x = -1 \) is invalid.