If log₂4`, `log_√28 and` log<sub>3</sub> 9<sup>k-1</sup>`

are consecutive terms of GP, then the number of integers that satisfy the system of inequalities x^2-x>6 and |x| < k^2 is

Option a 193
Option b 194
Option c 195
Option d 196

Options

A. 193
B. 194
C. 195
D. 196

Answer

To solve the problem step by step, we will first analyze the logarithmic expressions given in the question and then solve the inequalities.

### Step 1: Calculate \( \log_2 4 \)
We know that:
\[ \log_2 4 = \log_2 (2^2) = 2 \cdot \log_2 2 = 2 \cdot 1 = 2 \]
Thus, \( a = 2 \).

**Hint:** Recall that \( \log_b (a^n) = n \cdot \log_b a \).

### Step 2: Calculate \( \log_{\sqrt{2}} 8 \)
Using the change of base formula:
\[ \log_{\sqrt{2}} 8 = \log_{2^{1/2}} 8 = \frac{1}{\frac{1}{2}} \log_2 8 = 2 \log_2 8 \]
Now, since \( 8 = 2^3 \):
\[ \log_2 8 = 3 \Rightarrow \log_{\sqrt{2}} 8 = 2 \cdot 3 = 6 \]
Thus, \( b = 6 \).

**Hint:** Remember that \( \log_{b^m} a = \frac{1}{m} \log_b a \).

### Step 3: Calculate \( \log_3 9^{k-1} \)
We can express \( 9 \) as \( 3^2 \):
\[ \log_3 9^{k-1} = (k-1) \log_3 9 = (k-1) \cdot 2 = 2(k-1) \]
Thus, \( c = 2(k-1) \).

**Hint:** Use the property \( \log_b (a^n) = n \cdot \log_b a \).

### Step 4: Set up the GP condition
Since \( a, b, c \) are in GP, we have:
\[ b^2 = ac \]
Substituting the values:
\[ 6^2 = 2 \cdot 2(k-1) \]
This simplifies to:
\[ 36 = 4(k-1) \]
Dividing both sides by 4:
\[ 9 = k - 1 \Rightarrow k = 10 \]

**Hint:** Recall the condition for terms to be in GP: \( b^2 = ac \).

### Step 5: Solve the inequalities
We need to solve the inequalities \( x^2 - x > 6 \) and \( |x| < k^2 \).

1. **First inequality:**
\[ x^2 - x - 6 > 0 \]
Factoring gives:
\[ (x - 3)(x + 2) > 0 \]
The critical points are \( x = 3 \) and \( x = -2 \). Testing intervals:
- For \( x < -2 \), both factors are negative: \( > 0 \)
- For \( -2 < x < 3 \), one factor is negative and one is positive: \( < 0 \)
- For \( x > 3 \), both factors are positive: \( > 0 \)

Thus, the solution is:
\[ x < -2 \quad \text{or} \quad x > 3 \]

2. **Second inequality:**
\[ |x| < k^2 = 100 \Rightarrow -100 < x < 100 \]

**Hint:** Solve each inequality separately and then find the intersection.

### Step 6: Find the common solution
The combined solution from both inequalities is:
\[ (-100, -2) \cup (3, 100) \]

### Step 7: Count the integers in the intervals
1. **For the interval \((-100, -2)\)**:
- The integers are: \(-99, -98, \ldots, -3\) (total of 97 integers).

2. **For the interval \((3, 100)\)**:
- The integers are: \(4, 5, \ldots, 99\) (total of 96 integers).

Adding these gives:
\[ 97 + 96 = 193 \]

Thus, the total number of integers that satisfy the system of inequalities is **193**.

**Final Answer:** Option A: 193