Solution set of the in equality `log_(10^(2)) x-3(log_(10)x)( log_(10)(x-2))+2 log_(10^(2))(x-2) lt 0`, is :

Options

A. `(0,4)`
B. `(-oo,1)`
C. `(4,oo)`
D. `(2, 4)`

Answer

To solve the inequality \( \log_{10^2} x - 3(\log_{10} x)(\log_{10}(x-2)) + 2 \log_{10^2}(x-2) < 0 \), we will follow these steps:

### Step 1: Rewrite the logarithms
Using the property of logarithms, we can rewrite the logarithms with base \(10^2\):
\[ \log_{10^2} x = \frac{1}{2} \log_{10} x \]
\[ \log_{10^2}(x-2) = \frac{1}{2} \log_{10}(x-2) \]
Substituting these into the inequality gives:
\[ \frac{1}{2} \log_{10} x - 3(\log_{10} x)(\log_{10}(x-2)) + 2 \cdot \frac{1}{2} \log_{10}(x-2) < 0 \]
This simplifies to:
\[ \frac{1}{2} \log_{10} x - 3(\log_{10} x)(\log_{10}(x-2)) + \log_{10}(x-2) < 0 \]

### Step 2: Multiply through by 2
To eliminate the fraction, multiply the entire inequality by 2 (which does not change the direction of the inequality):
\[ \log_{10} x - 6(\log_{10} x)(\log_{10}(x-2)) + 2 \log_{10}(x-2) < 0 \]

### Step 3: Combine logarithms
We can combine the logarithmic terms:
\[ \log_{10} x + 2 \log_{10}(x-2) < 6(\log_{10} x)(\log_{10}(x-2)) \]
This can be rewritten as:
\[ \log_{10}(x) + \log_{10}((x-2)^2) < 6(\log_{10} x)(\log_{10}(x-2)) \]
Using the property of logarithms, we can combine the left side:
\[ \log_{10}(x(x-2)^2) < 6(\log_{10} x)(\log_{10}(x-2)) \]

### Step 4: Define new variables
Let \( a = \log_{10} x \) and \( b = \log_{10}(x-2) \). The inequality becomes:
\[ \log_{10}(10^a \cdot 10^{2b}) < 6ab \]
This simplifies to:
\[ a + 2b < 6ab \]

### Step 5: Rearranging the inequality
Rearranging gives:
\[ 6ab - a - 2b > 0 \]
Factoring out gives:
\[ a(6b - 1) - 2b > 0 \]

### Step 6: Analyze critical points
To find the critical points, we set:
\[ a(6b - 1) = 2b \]
This gives us two cases to analyze:
1. \( a = 0 \) which implies \( x = 1 \)
2. \( 6b - 1 = 0 \) which implies \( b = \frac{1}{6} \) or \( x - 2 = 10^{\frac{1}{6}} \)

### Step 7: Determine the intervals
We need to check the intervals defined by \( x = 1 \) and \( x = 2 + 10^{\frac{1}{6}} \) to determine where the inequality holds.

### Step 8: Check the values
- For \( x < 2 \): The logarithm is not defined.
- For \( 2 < x < 3 \): Check \( x = 3 \) in the original inequality.
- For \( x > 3 \): Check if the inequality holds.

### Conclusion
After checking the values, we find that the solution set of the inequality is:
\[ x \in (2, 3) \cup (3, \infty) \]