Which of the following is not the solution of `(log)_3(x^2-2)<(log)_3(3/2|x|-1)` is `(sqrt(2),2)` (b) `(-2,-sqrt(2))` `(-sqrt(2),2` (d) none of these

Options

A. `(-sqrt2,2)`
B. `(-2, -sqrt2)`
C. ` (-sqrt2, 2)`
D. none of these

Explanation

To solve the inequality \( \log_3(x^2 - 2) < \log_3\left(\frac{3}{2}|x| - 1\right) \), we will follow these steps:

### Step 1: Determine the domain of the logarithmic functions
The logarithmic functions are defined only when their arguments are positive. Therefore, we need to find the conditions under which \( x^2 - 2 > 0 \) and \( \frac{3}{2}|x| - 1 > 0 \).

1. **For \( x^2 - 2 > 0 \)**:
\[ x^2 > 2 \implies |x| > \sqrt{2} \implies x < -\sqrt{2} \text{ or } x > \sqrt{2} \]

2. **For \( \frac{3}{2}|x| - 1 > 0 \)**:
\[ \frac{3}{2}|x| > 1 \implies |x| > \frac{2}{3} \implies x < -\frac{2}{3} \text{ or } x > \frac{2}{3} \]

### Step 2: Combine the conditions
From the above inequalities, we have:
- From \( x^2 - 2 > 0 \): \( x < -\sqrt{2} \) or \( x > \sqrt{2} \)
- From \( \frac{3}{2}|x| - 1 > 0 \): \( x < -\frac{2}{3} \) or \( x > \frac{2}{3} \)

Now we need to find the intersection of these conditions:
- For \( x < -\sqrt{2} \): This is already less than \(-\frac{2}{3}\).
- For \( x > \sqrt{2} \): This is already greater than \(\frac{2}{3}\).

Thus, the combined conditions for the domain are:
- \( x < -\sqrt{2} \) or \( x > \sqrt{2} \)

### Step 3: Solve the inequality
Since the base of the logarithm (3) is greater than 1, the inequality sign remains the same when we remove the logarithm:
\[ x^2 - 2 < \frac{3}{2}|x| - 1 \]

Rearranging gives:
\[ x^2 - \frac{3}{2}|x| + 1 < 0 \]

### Step 4: Analyze the quadratic inequality
Let \( y = |x| \). The inequality becomes:
\[ y^2 - \frac{3}{2}y + 1 < 0 \]

To find the roots of the quadratic equation \( y^2 - \frac{3}{2}y + 1 = 0 \), we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{\frac{3}{2} \pm \sqrt{\left(\frac{3}{2}\right)^2 - 4 \cdot 1}}{2} \]
\[ = \frac{\frac{3}{2} \pm \sqrt{\frac{9}{4} - 4}}{2} = \frac{\frac{3}{2} \pm \sqrt{\frac{9 - 16}{4}}}{2} = \frac{\frac{3}{2} \pm \sqrt{-\frac{7}{4}}}{2} \]

Since the discriminant is negative, the quadratic has no real roots, meaning it does not cross the x-axis and is always positive or always negative.

### Step 5: Determine the sign of the quadratic
To determine the sign of the quadratic, we can evaluate it at a test point. For \( y = 0 \):
\[ 0^2 - \frac{3}{2}(0) + 1 = 1 > 0 \]
Thus, \( y^2 - \frac{3}{2}y + 1 > 0 \) for all \( y \).

### Conclusion
The inequality \( y^2 - \frac{3}{2}y + 1 < 0 \) has no solutions. Therefore, the solution to the original inequality is determined solely by the domain conditions.

### Final Answer
The solutions are:
- \( x < -\sqrt{2} \) or \( x > \sqrt{2} \)

Thus, the option that is **not** a solution is:
(b) \((-2, -\sqrt{2})\)