What is the least integral value of `2log_(10) x - log_(x)` (0.01) ?

Options

A. 0
B. 2
C. 4
D. 3

Answer

To find the least integral value of the expression \( 2\log_{10} x - \log_{x}(0.01) \), we will simplify and analyze the expression step by step.

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

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

### Step 3: Let \( y = \log_{10}(x) \)
Let \( y = \log_{10}(x) \). Then the expression simplifies to:
\[ 2y + \frac{2}{y} \]

### Step 4: Find the minimum value of the expression
To find the minimum value of \( 2y + \frac{2}{y} \), we can take the derivative and set it to zero:
\[ \frac{d}{dy}(2y + \frac{2}{y}) = 2 - \frac{2}{y^2} \]
Setting the derivative to zero:
\[ 2 - \frac{2}{y^2} = 0 \implies \frac{2}{y^2} = 2 \implies y^2 = 1 \implies y = 1 \text{ (since } y > 0\text{)} \]

### Step 5: Evaluate the second derivative
To confirm that this point is a minimum, we can check the second derivative:
\[ \frac{d^2}{dy^2}(2y + \frac{2}{y}) = \frac{4}{y^3} \]
Since \( y > 0 \), the second derivative is positive, confirming a local minimum.

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

### Step 7: Conclusion
Thus, the least integral value of \( 2\log_{10} x - \log_{x}(0.01) \) is:
\[ \boxed{4} \]