Question

Find the number of real values of x satisfying the equation:
\[ \log_{2}(x^2 - 5x + 6) + \log_{1/2}(x - 2) = 3 \]

Hint:

Before combining anything, work out exactly which x-values make both log arguments positive — this domain restriction is what rules out any extra candidate later. Also remember that a log with base $1/2$ is just the negative of the same log with base $2$, so either one can be rewritten in terms of the other.

Answer 1

Correct Answer: 1

Approach: Fix the domain first (logs of odd things bite you here), then convert the base-\(1/2\) log to a negative base-\(2\) log so the whole left side collapses to a single log.

Step 1 (domain): Need \(x^2 - 5x + 6 > 0\), i.e. \((x-2)(x-3) > 0\), giving \(x < 2\) or \(x > 3\). Also need \(x - 2 > 0\), i.e. \(x > 2\). The intersection is \(x > 3\).

Step 2: Use \(\log_{1/2} a = -\log_2 a\). The equation becomes \[ \log_2(x^2 - 5x + 6) - \log_2(x-2) = 3. \]
Step 3: Combine and factor the numerator \(x^2-5x+6 = (x-2)(x-3)\): \[ \log_2\!\left(\frac{(x-2)(x-3)}{x-2}\right) = \log_2(x-3) = 3. \] Cancelling \((x-2)\) is legal since \(x>3\).

Step 4: So \(x - 3 = 2^3 = 8 \Rightarrow x = 11\), which satisfies \(x>3\).

Step 5: Exactly one valid value of \(x\) survives.

Answer: Number of real solutions \(= \boxed{1}\).

Answer 2

Approach: Rewrite \(\log_{1/2}\) in base \(2\) so both terms combine into a single logarithm, then solve the resulting equation while tracking the domain throughout.

Since \(\log_{1/2}(x-2)=-\log_{2}(x-2)\), the equation becomes \[ \log_{2}(x^2-5x+6)-\log_{2}(x-2)=3 \implies \log_{2}\!\left(\frac{x^2-5x+6}{x-2}\right)=3. \]
Factor the numerator: \(x^2-5x+6=(x-2)(x-3)\), so for \(x\neq2\) the fraction simplifies to \(x-3\). The equation becomes \(\log_{2}(x-3)=3 \implies x-3=8 \implies x=11\).

Domain check: we need \(x^2-5x+6>0\) (i.e. \(x<2\) or \(x>3\)) and \(x-2>0\) (i.e. \(x>2\)); together these force \(x>3\), and \(x=11\) satisfies this.

Since the simplification produces exactly one candidate and it survives the domain check, there is \[ \boxed{\text{exactly 1 real value of } x} \]

Answer 3

Concept:

• $\log_2(a) = -\log_{1/2}(a)$, since $2$ and $1/2$ are reciprocal bases — this lets an entire equation be rewritten in base $1/2$ instead of base $2$.
• $\log_{1/2}(a) - \log_{1/2}(b) = \log_{1/2}(a/b)$: the quotient rule for logarithms holds in any base, including a fractional one.
• $\log_{1/2}(N) = k$ means $(1/2)^k = N$, so a fractional base still converts to an ordinary exponential equation once the log is isolated.

Step 1: Fix the domain first.
Need $x^2 - 5x + 6 = (x-2)(x-3) > 0$, giving $x < 2$ or $x > 3$. Also need $x - 2 > 0$, giving $x > 2$. Combined domain: $x > 3$.

Step 2: Convert the base-$2$ term into base $1/2$.
$\log_2(x^2-5x+6) = -\log_{1/2}(x^2-5x+6)$. Substituting into the original equation:
$-\log_{1/2}(x^2-5x+6) + \log_{1/2}(x-2) = 3$

Step 3: Combine into a single base-$1/2$ log and simplify.
$\log_{1/2}(x-2) - \log_{1/2}(x^2-5x+6) = \log_{1/2}\left(\dfrac{x-2}{(x-2)(x-3)}\right) = \log_{1/2}\left(\dfrac{1}{x-3}\right)$
(cancelling $x-2$ is valid since $x > 3$ means $x-2 \neq 0$), so $\log_{1/2}\left(\dfrac{1}{x-3}\right) = 3$.

Step 4: Convert to exponential form and solve.
$\dfrac{1}{x-3} = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8} \Rightarrow x - 3 = 8 \Rightarrow x = 11$, which lies in the domain $x>3$.

Final Answer: Number of real solutions $= 1$.