Solve the following inequation .
(vii) `log_(10)x+2lelog_10^2x`

Answer

To solve the inequation \( \log_{10} x + 2 \leq \log_{10}^2 x \), we can follow these steps:

### Step 1: Rewrite the Inequation
Start by rewriting the inequation:
\[ \log_{10} x + 2 \leq \log_{10}^2 x \]
This can be rearranged to:
\[ \log_{10}^2 x - \log_{10} x - 2 \geq 0 \]

### Step 2: Let \( y = \log_{10} x \)
Let \( y = \log_{10} x \). Then, the inequation becomes:
\[ y^2 - y - 2 \geq 0 \]

### Step 3: Factor the Quadratic
Next, we factor the quadratic expression:
\[ y^2 - y - 2 = (y - 2)(y + 1) \]
Thus, we have:
\[ (y - 2)(y + 1) \geq 0 \]

### Step 4: Find the Critical Points
The critical points are found by setting each factor to zero:
\[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \]
\[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \]

### Step 5: Test Intervals
Now, we will test the intervals determined by the critical points \( y = -1 \) and \( y = 2 \):
1. **Interval \( (-\infty, -1) \)**: Choose \( y = -2 \):
\[ (-2 - 2)(-2 + 1) = (-4)(-1) = 4 \geq 0 \quad \text{(True)} \]
2. **Interval \( (-1, 2) \)**: Choose \( y = 0 \):
\[ (0 - 2)(0 + 1) = (-2)(1) = -2 < 0 \quad \text{(False)} \]
3. **Interval \( (2, \infty) \)**: Choose \( y = 3 \):
\[ (3 - 2)(3 + 1) = (1)(4) = 4 \geq 0 \quad \text{(True)} \]

### Step 6: Combine the Results
The solution to the inequality is:
\[ y \in (-\infty, -1] \cup [2, \infty) \]

### Step 7: Convert Back to \( x \)
Recall that \( y = \log_{10} x \). Therefore, we convert back:
1. For \( y \leq -1 \):
\[ \log_{10} x \leq -1 \quad \Rightarrow \quad x \leq 10^{-1} = 0.1 \]
2. For \( y \geq 2 \):
\[ \log_{10} x \geq 2 \quad \Rightarrow \quad x \geq 10^2 = 100 \]

### Final Solution
Thus, the solution to the inequation is:
\[ x \in (0, 0.1] \cup [100, \infty) \]