If `(log_(3)x)^(2)+log_(3)x lt2`, then which one of the following is correct?

Options

A. `0 lt x lt 1/9`
B. ` 1/9 le x le 3`
C. `3 lt x lt oo`
D. ` 1/9 le x le 3`

Explanation

To solve the inequality \((\log_{3} x)^{2} + \log_{3} x < 2\), we will follow these steps:

### Step 1: Substitute \(y\) for \(\log_{3} x\)
Let \(y = \log_{3} x\). Then, the inequality becomes:
\[ y^2 + y < 2 \]

### Step 2: Rearrange the inequality
Rearranging gives us:
\[ y^2 + y - 2 < 0 \]

### Step 3: Factor the quadratic expression
We need to factor the quadratic expression \(y^2 + y - 2\). We look for two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\). Thus, we can factor it as:
\[ (y + 2)(y - 1) < 0 \]

### Step 4: Determine the critical points
The critical points from the factors are:
\[ y + 2 = 0 \quad \Rightarrow \quad y = -2 \]
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]

### Step 5: Test intervals
We will test the intervals determined by the critical points \(-2\) and \(1\):
- **Interval 1**: \(y < -2\) (e.g., \(y = -3\))
\(((-3) + 2)((-3) - 1) = (-1)(-4) > 0\)

- **Interval 2**: \(-2 < y < 1\) (e.g., \(y = 0\))
\(((0) + 2)((0) - 1) = (2)(-1) < 0\)

- **Interval 3**: \(y > 1\) (e.g., \(y = 2\))
\(((2) + 2)((2) - 1) = (4)(1) > 0\)

### Step 6: Write the solution in terms of \(y\)
The inequality \((y + 2)(y - 1) < 0\) holds true in the interval:
\[ -2 < y < 1 \]

### Step 7: Substitute back for \(x\)
Now, substituting back for \(y\):
\[ -2 < \log_{3} x < 1 \]

### Step 8: Convert the logarithmic inequalities to exponential form
1. From \(-2 < \log_{3} x\):
\[ x > 3^{-2} = \frac{1}{9} \]

2. From \(\log_{3} x < 1\):
\[ x < 3^{1} = 3 \]

### Final Solution
Thus, the solution for \(x\) is:
\[ \frac{1}{9} < x < 3 \]