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:

For logarithmic equations, always find the domain of the variable first. This helps in rejecting extraneous solutions obtained during the algebraic manipulation.

Answer 1

Correct Answer: 1

Step 1: Identify the domain restrictions. For each logarithmic term to be defined, its argument must be positive. From \[ \log_2(x^2 - 5x + 6), \] we get \[ x^2 - 5x + 6 > 0 \Rightarrow (x-2)(x-3) > 0, \] which holds for \[ x < 2 \quad \text{or} \quad x > 3. \] From \[ \log_{1/2}(x-2), \] we require \[ x - 2 > 0 \Rightarrow x > 2. \] Combining both conditions, the common domain is \[ x > 3. \] Step 2: Rewrite the equation using base 2. Using the property \(\log_{1/2} a = -\log_2 a\), the given equation becomes \[ \log_2(x^2 - 5x + 6) - \log_2(x-2) = 3. \] Applying the logarithmic subtraction rule, \[ \log_2\!\left(\frac{x^2 - 5x + 6}{x-2}\right) = 3. \] Step 3: Simplify the expression. Factor the quadratic in the numerator: \[ x^2 - 5x + 6 = (x-2)(x-3). \] Thus, \[ \log_2\!\left(\frac{(x-2)(x-3)}{x-2}\right) = 3. \] Since \(x > 3\), cancellation is valid, giving \[ \log_2(x-3) = 3. \] Step 4: Solve for \(x\). Converting to exponential form, \[ x - 3 = 2^3 = 8, \] so \[ x = 11. \] Step 5: Conclude the number of solutions. The obtained value satisfies the domain condition \(x > 3\). Therefore, there is exactly one real solution to the equation.

Answer 2

Step 1: Understanding the Question:
The question asks for the number of real solutions to a logarithmic equation. To solve this, we must first determine the valid domain for 'x' by ensuring the arguments of the logarithms are positive, and then solve the equation using logarithmic properties.
Step 2: Key Formula or Approach:
We will use the following properties of logarithms:
1. Domain of Logarithm: For \(\log_{b}(a)\) to be defined, we must have \(a>0\).
2. Change of Base Formula: \(\log_{b^n}(a) = \frac{1}{n} \log_{b}(a)\). A special case is \(\log_{1/b}(a) = \log_{b^{-1}}(a) = -\log_{b}(a)\).
3. Logarithm of a Quotient: \(\log_{b}(a) - \log_{b}(c) = \log_{b}\left(\frac{a}{c}\right)\).
4. Logarithmic to Exponential Form: If \(\log_{b}(a) = c\), then \(a = b^c\).
Step 3: Detailed Explanation:
Part A: Finding the Domain
The arguments of the logarithms must be strictly positive.
1. From \(\log_{2}(x^2 - 5x + 6)\), we have:
\[ x^2 - 5x + 6>0 \] \[ (x - 2)(x - 3)>0 \] This inequality is true when \(x<2\) or \(x>3\).
2. From \(\log_{1/2}(x - 2)\), we have:
\[ x - 2>0 \] \[ x>2 \] To satisfy both conditions, we take the intersection of the two solution sets: (\(x<2\) or \(x>3\)) AND (\(x>2\)). The common domain is \(x>3\). Any valid solution must satisfy this condition.
Part B: Solving the Equation
First, we simplify the second term using the change of base property:
\[ \log_{1/2}(x - 2) = \log_{2^{-1}}(x - 2) = -1 \cdot \log_{2}(x - 2) = -\log_{2}(x - 2) \] Substitute this back into the original equation:
\[ \log_{2}(x^2 - 5x + 6) - \log_{2}(x - 2) = 3 \] Now, use the quotient rule for logarithms:
\[ \log_{2}\left(\frac{x^2 - 5x + 6}{x - 2}\right) = 3 \] Factor the numerator:
\[ \log_{2}\left(\frac{(x - 2)(x - 3)}{x - 2}\right) = 3 \] Since our domain is \(x>3\), \(x - 2 \neq 0\), so we can cancel the term:
\[ \log_{2}(x - 3) = 3 \] Convert this to exponential form:
\[ x - 3 = 2^3 \] \[ x - 3 = 8 \] \[ x = 11 \] Part C: Checking the Solution
The solution we found is \(x = 11\). We must check if it lies in the valid domain (\(x>3\)).
Since \(11>3\), the solution is valid.
Step 4: Final Answer:
There is only one real value, \(x = 11\), that satisfies the equation. Therefore, the number of solutions is 1.