The set of all `x` satisfying `3^(2x)-3x^(x)-6gt0` is given by

Options

A. `0ltxlt1`
B. `xgt1`
C. `xgt3^(-2)`
D. none of these

Answer

To solve the inequality \( 3^{2x} - 3x^x - 6 > 0 \), we can follow these steps:

### Step 1: Substitute \( t = 3^x \)
We start by substituting \( t = 3^x \). This gives us:
\[ 3^{2x} = (3^x)^2 = t^2 \]
Thus, the inequality becomes:
\[ t^2 - 3t - 6 > 0 \]

### Step 2: Factor the quadratic inequality
Next, we need to factor the quadratic expression \( t^2 - 3t - 6 \). We can rewrite it as:
\[ t^2 - 3t + 2t - 6 > 0 \]
Grouping the terms, we have:
\[ t(t - 3) + 2(t - 3) > 0 \]
Factoring out \( (t - 3) \):
\[ (t - 3)(t + 2) > 0 \]

### Step 3: Find the critical points
The critical points of the inequality are found by setting each factor to zero:
\[ t - 3 = 0 \quad \Rightarrow \quad t = 3 \]
\[ t + 2 = 0 \quad \Rightarrow \quad t = -2 \]

### Step 4: Analyze the sign of the factors
We will analyze the sign of \( (t - 3)(t + 2) \) on the number line. The critical points divide the number line into intervals:
1. \( (-\infty, -2) \)
2. \( (-2, 3) \)
3. \( (3, \infty) \)

We will test a point from each interval to determine where the product is positive.

- For \( t < -2 \) (e.g., \( t = -3 \)):
\[ (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 \]
- For \( -2 < t < 3 \) (e.g., \( t = 0 \)):
\[ (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 \]
- For \( t > 3 \) (e.g., \( t = 4 \)):
\[ (4 - 3)(4 + 2) = (1)(6) = 6 > 0 \]

### Step 5: Determine the solution set
The product \( (t - 3)(t + 2) > 0 \) is satisfied in the intervals:
1. \( t < -2 \) (not possible since \( t = 3^x \) is always positive)
2. \( t > 3 \)

Thus, we only consider \( t > 3 \).

### Step 6: Convert back to \( x \)
Since \( t = 3^x \), we have:
\[ 3^x > 3 \]
Taking logarithm base 3 on both sides:
\[ x > 1 \]

### Final Answer
The set of all \( x \) satisfying the inequality \( 3^{2x} - 3x^x - 6 > 0 \) is:
\[ \boxed{(1, \infty)} \]