Question

Find the real value(s) of x that satisfy the equation:
\[ \log_{2}(x^2 - 5x + 6) + \log_{1/2}(x - 2) = 3 \]

Hint:

The first and most critical step in solving logarithmic equations is to determine the domain of the variable. This helps you to immediately discard any extraneous solutions you might find during your calculations.

Answer 1

Correct Answer: 11

Approach: Same machinery as any mixed-base log equation — nail the domain, flip the base-\(1/2\) log into a negative base-\(2\) log, then collapse to one log and solve.

Step 1 (domain): \(x^2 - 5x + 6 > 0 \Rightarrow (x-2)(x-3) > 0 \Rightarrow x < 2\) or \(x > 3\); and \(x - 2 > 0 \Rightarrow x > 2\). Intersection: \(x > 3\).

Step 2: Since \(\log_{1/2}(x-2) = -\log_2(x-2)\), the equation becomes \[ \log_2(x^2 - 5x + 6) - \log_2(x-2) = 3. \]
Step 3: Factor \(x^2 - 5x + 6 = (x-2)(x-3)\) and cancel (legal as \(x>3\)): \[ \log_2\!\left(\frac{(x-2)(x-3)}{x-2}\right) = \log_2(x-3) = 3. \]
Step 4: \(x - 3 = 2^3 = 8 \Rightarrow x = 11\), and \(11 > 3\) is within the domain.

Answer: \(\boxed{x = 11}\).

Answer 2

Approach: Convert the log equation into an exponential one right away, then substitute \(y=x-2\) to simplify the algebra.

Combine the logarithms (using \(\log_{1/2}(x-2)=-\log_2(x-2)\)): \[ \log_2\left(\frac{x^2-5x+6}{x-2}\right)=3 \implies \frac{x^2-5x+6}{x-2}=2^3=8. \]
Since \(x^2-5x+6=(x-2)(x-3)\), letting \(y=x-2\) turns the left side into \(\dfrac{y(y-1)}{y}=y-1\) (valid since \(y\neq0\) in the domain). So \[ y-1=8 \implies y=9 \implies x=y+2=11. \]
Check the domain: \(x-2=9>0\) and \(x^2-5x+6=72>0\), both satisfied.

\[ \boxed{x=11} \]