If `(log_(3)x)^(2)+log_(3)xlt2`, then which one of the following is correct ?

Options

A. `0ltxlt1/9`
B. `1/9ltxlt3`
C. `3ltxltoo`
D. `1/9lexle3`

Answer

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

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

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

### Step 3: Factor the quadratic expression
Now we need to factor the quadratic expression \(t^{2} + t - 2\). We can rewrite it as:
\[ t^{2} + 2t - t - 2 < 0 \]
Grouping the terms, we have:
\[ t(t + 2) - 1(t + 2) < 0 \]
Factoring out \((t + 2)\):
\[ (t + 2)(t - 1) < 0 \]

### Step 4: Find the critical points
The critical points occur when each factor equals zero:
1. \(t + 2 = 0 \Rightarrow t = -2\)
2. \(t - 1 = 0 \Rightarrow t = 1\)

### Step 5: Test intervals on the number line
We will test the intervals determined by the critical points \(-2\) and \(1\):
- Interval 1: \(t < -2\)
- Interval 2: \(-2 < t < 1\)
- Interval 3: \(t > 1\)

Choose test points from each interval:
1. For \(t = -3\) (Interval 1): \((-3 + 2)(-3 - 1) = (-1)(-4) > 0\) (positive)
2. For \(t = 0\) (Interval 2): \((0 + 2)(0 - 1) = (2)(-1) < 0\) (negative)
3. For \(t = 2\) (Interval 3): \((2 + 2)(2 - 1) = (4)(1) > 0\) (positive)

### Step 6: Write the solution for \(t\)
The inequality \((t + 2)(t - 1) < 0\) holds in the interval:
\[ -2 < t < 1 \]

### Step 7: Substitute back for \(x\)
Since \(t = \log_{3} x\), we have:
\[ -2 < \log_{3} x < 1 \]

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

### Step 9: Combine the inequalities
Combining these results gives us:
\[ \frac{1}{9} < x < 3 \]

### Conclusion
Thus, the correct answer is:
\[ \frac{1}{9} < x < 3 \]