Solve the following inequation .
(xv) `x^((log_10x)^2-3log_10x+1)gt1000`

Answer

To solve the inequation \( x^{(\log_{10} x)^2 - 3 \log_{10} x + 1} > 1000 \), we can follow these steps:

### Step 1: Rewrite the Inequation
We start with the given inequation:
\[ x^{(\log_{10} x)^2 - 3 \log_{10} x + 1} > 1000 \]
We can express \( 1000 \) as \( 10^3 \):
\[ x^{(\log_{10} x)^2 - 3 \log_{10} x + 1} > 10^3 \]

### Step 2: Take Logarithm
Taking logarithm base \( 10 \) on both sides:
\[ \log_{10} \left( x^{(\log_{10} x)^2 - 3 \log_{10} x + 1} \right) > \log_{10}(10^3) \]
Using the property of logarithms \( \log(a^b) = b \cdot \log(a) \):
\[ ((\log_{10} x)^2 - 3 \log_{10} x + 1) \cdot \log_{10} x > 3 \]

### Step 3: Substitute \( t \)
Let \( t = \log_{10} x \). Then we can rewrite the inequation as:
\[ (t^2 - 3t + 1) t > 3 \]
This simplifies to:
\[ t^3 - 3t^2 + t - 3 > 0 \]

### Step 4: Rearranging the Inequation
Rearranging gives us:
\[ t^3 - 3t^2 + t - 3 > 0 \]

### Step 5: Factor the Polynomial
We can factor the polynomial:
\[ t^3 - 3t^2 + t - 3 = (t - 3)(t^2 + 1) \]
Since \( t^2 + 1 \) is always positive for all real \( t \), we focus on the factor \( t - 3 \):
\[ (t - 3)(t^2 + 1) > 0 \]

### Step 6: Solve the Inequality
The inequality \( t - 3 > 0 \) implies:
\[ t > 3 \]
Substituting back for \( t \):
\[ \log_{10} x > 3 \]

### Step 7: Exponentiate to Solve for \( x \)
Exponentiating both sides gives:
\[ x > 10^3 \]
Thus, we find:
\[ x > 1000 \]

### Final Solution
The solution set is:
\[ x \in (1000, \infty) \]