The least value of expression `2log_(10)x-log_(x)(1//100)` for `xgt1` is ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the least value of the expression \( 2\log_{10}x - \log_{x}\left(\frac{1}{100}\right) \) for \( x > 1 \), we can follow these steps:

### Step-by-Step Solution

1. **Rewrite the logarithm**:
We know that \( \log_{x}\left(\frac{1}{100}\right) \) can be rewritten using the change of base formula:
\[ \log_{x}\left(\frac{1}{100}\right) = \frac{\log_{10}\left(\frac{1}{100}\right)}{\log_{10}x} \]
Since \( \frac{1}{100} = 10^{-2} \), we have:
\[ \log_{10}\left(\frac{1}{100}\right) = -2 \]
Therefore:
\[ \log_{x}\left(\frac{1}{100}\right) = \frac{-2}{\log_{10}x} \]

2. **Substitute back into the expression**:
Substitute this back into the original expression:
\[ 2\log_{10}x - \log_{x}\left(\frac{1}{100}\right) = 2\log_{10}x - \left(\frac{-2}{\log_{10}x}\right) \]
This simplifies to:
\[ 2\log_{10}x + \frac{2}{\log_{10}x} \]

3. **Let \( y = \log_{10}x \)**:
Since \( x > 1 \), \( y > 0 \). The expression now becomes:
\[ 2y + \frac{2}{y} \]

4. **Find the minimum value**:
To find the minimum value of \( 2y + \frac{2}{y} \), we can use calculus or the AM-GM inequality. Using AM-GM:
\[ \frac{2y + \frac{2}{y}}{2} \geq \sqrt{2y \cdot \frac{2}{y}} = 2 \]
Therefore:
\[ 2y + \frac{2}{y} \geq 4 \]

5. **Equality condition**:
The equality holds when \( 2y = \frac{2}{y} \), which leads to \( y^2 = 1 \) or \( y = 1 \). Since \( y = \log_{10}x \), this means \( x = 10 \).

6. **Conclusion**:
The least value of the expression \( 2\log_{10}x - \log_{x}\left(\frac{1}{100}\right) \) for \( x > 1 \) is:
\[ \boxed{4} \]