The least value of the expression `2log_(10)x-log_(x)0.01` for `xgt1` is …..

Answer

To find the least value of the expression \(2 \log_{10} x - \log_x 0.01\) for \(x > 1\), we can follow these steps:

### Step 1: Rewrite the logarithmic expression
We start with the expression:
\[ E = 2 \log_{10} x - \log_x 0.01 \]
Using the change of base formula, we can rewrite \(\log_x 0.01\):
\[ \log_x 0.01 = \frac{\log_{10} 0.01}{\log_{10} x} \]
Since \(0.01 = 10^{-2}\), we have:
\[ \log_{10} 0.01 = -2 \]
Thus, we can rewrite \(\log_x 0.01\) as:
\[ \log_x 0.01 = \frac{-2}{\log_{10} x} \]

### Step 2: Substitute back into the expression
Now substituting this back into our expression \(E\):
\[ E = 2 \log_{10} x - \left(\frac{-2}{\log_{10} x}\right) \]
This simplifies to:
\[ E = 2 \log_{10} x + \frac{2}{\log_{10} x} \]

### Step 3: Let \(y = \log_{10} x\)
Let \(y = \log_{10} x\). Since \(x > 1\), we have \(y > 0\). Now the expression becomes:
\[ E = 2y + \frac{2}{y} \]

### Step 4: Find the minimum value of \(E\)
To find the minimum value of \(E\), we can take the derivative and set it to zero:
\[ \frac{dE}{dy} = 2 - \frac{2}{y^2} \]
Setting the derivative equal to zero:
\[ 2 - \frac{2}{y^2} = 0 \]
This leads to:
\[ \frac{2}{y^2} = 2 \quad \Rightarrow \quad y^2 = 1 \quad \Rightarrow \quad y = 1 \quad (\text{since } y > 0) \]

### Step 5: Verify it's a minimum
To confirm that this is a minimum, we can check the second derivative:
\[ \frac{d^2E}{dy^2} = \frac{4}{y^3} \]
Since \(y > 0\), \(\frac{d^2E}{dy^2} > 0\), indicating that \(E\) is concave up at \(y = 1\) and thus a minimum.

### Step 6: Calculate the minimum value
Substituting \(y = 1\) back into the expression for \(E\):
\[ E = 2(1) + \frac{2}{1} = 2 + 2 = 4 \]

### Final Answer
The least value of the expression \(2 \log_{10} x - \log_x 0.01\) for \(x > 1\) is:
\[ \boxed{4} \]