Question

The set of all values of x satisfying the inequality \( \log_{(x+\frac{1}{2})} \left[ \log_2 \left( \frac{x-1}{x+2} \right) \right]>0 \) is

• A. \( (-5, -2) \)
• B. \( (2, 5) \)
• C. Null set
• D. \( (5, \infty) \)

Hint:

With complex logarithmic inequalities, always start by finding the domain. In many cases, the domain constraints might be contradictory, leading to a null set solution without needing to solve the main inequality itself. This can save a significant amount of time.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Solving logarithmic inequalities requires careful consideration of the domain of the functions involved. The solution must satisfy both the domain conditions and the inequality itself.
Step 2: Key Formula or Approach:
For an expression \( \log_b(A) \) to be defined, we must have:

Argument is positive: \( A>0 \)
Base is positive: \( b>0 \)
Base is not equal to 1: \( b \neq 1 \)
For the inequality \( \log_b(A)>0 \), we have two cases:

If \( b>1 \), then \( A>1 \).
If \( 0<b<1 \), then \( 0<A<1 \).
Step 3: Detailed Explanation:
Let's first establish the domain for the given expression. Let the base be \( b = x + \frac{1}{2} \) and the argument be \( A = \log_2 \left( \frac{x-1}{x+2} \right) \). Domain Conditions:

For the inner logarithm, \( \log_2(\dots) \): Its argument must be positive: \[ \frac{x-1}{x+2}>0 \] This is true when \( x>1 \) or \( x<-2 \). (Condition I)
For the outer logarithm, \( \log_{x+1/2}(\dots) \):

Its base must be positive and not 1: \[ x + \frac{1}{2}>0 \implies x>-\frac{1}{2} \] \[ x + \frac{1}{2} \neq 1 \implies x \neq \frac{1}{2} \] (Condition II)
Its argument \( A = \log_2 \left( \frac{x-1}{x+2} \right) \) must be positive: \[ \log_2 \left( \frac{x-1}{x+2} \right)>0 \] Since the base is 2 (which is>1), this implies: \[ \frac{x-1}{x+2}>2^0 = 1 \] \[ \frac{x-1}{x+2} - 1>0 \] \[ \frac{(x-1) - (x+2)}{x+2}>0 \] \[ \frac{-3}{x+2}>0 \] This inequality holds only if the denominator is negative, so: \[ x+2<0 \implies x<-2 \] (Condition III)

Combining the conditions: We need to find the values of x that satisfy all the domain conditions simultaneously. From Condition II, we must have \( x>-\frac{1}{2} \). From Condition III, we must have \( x<-2 \). There is no real number x that can be both greater than \(-\frac{1}{2}\) and less than \(-2\) at the same time. The intersection of these conditions is an empty set. Since the domain of the inequality is the null set, there are no values of x for which the inequality is even defined.
Step 4: Final Answer:
The set of all values of x satisfying the inequality is the Null set.