Find the least value of` log_2x-log_x(0.125)`for `xgt1` .

Answer

To find the least value of \( \log_2 x - \log_x (0.125) \) for \( x > 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \log_2 x - \log_x (0.125) \]
We can rewrite \( 0.125 \) as \( \frac{1}{8} \) or \( 8^{-1} \). Thus, we have:
\[ \log_x (0.125) = \log_x (8^{-1}) = -\log_x (8) \]
So the expression becomes:
\[ \log_2 x + \log_x (8) \]

### Step 2: Use the change of base formula
Using the change of base formula for logarithms, we can express \( \log_x (8) \) as:
\[ \log_x (8) = \frac{\log_2 (8)}{\log_2 (x)} \]
Since \( 8 = 2^3 \), we know that \( \log_2 (8) = 3 \). Therefore:
\[ \log_x (8) = \frac{3}{\log_2 (x)} \]

### Step 3: Substitute back into the expression
Now substituting this back into our expression gives:
\[ \log_2 x + \frac{3}{\log_2 x} \]
Let \( y = \log_2 x \). Then we can rewrite the expression as:
\[ y + \frac{3}{y} \]

### Step 4: Find the minimum value
To find the minimum value of \( y + \frac{3}{y} \), we can use calculus. We first find the derivative:
\[ f(y) = y + \frac{3}{y} \]
The derivative \( f'(y) \) is:
\[ f'(y) = 1 - \frac{3}{y^2} \]
Setting the derivative to zero to find critical points:
\[ 1 - \frac{3}{y^2} = 0 \implies \frac{3}{y^2} = 1 \implies y^2 = 3 \implies y = \sqrt{3} \]
Since \( x > 1 \), we have \( y > 0 \).

### Step 5: Verify the minimum
To confirm that this is a minimum, we check the second derivative:
\[ f''(y) = \frac{6}{y^3} \]
Since \( f''(y) > 0 \) for \( y > 0 \), the function is concave up at \( y = \sqrt{3} \), confirming a local minimum.

### Step 6: Calculate the minimum value
Now substituting \( y = \sqrt{3} \) back into the expression:
\[ f(\sqrt{3}) = \sqrt{3} + \frac{3}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \]

Thus, the least value of \( \log_2 x - \log_x (0.125) \) for \( x > 1 \) is:
\[ \boxed{2\sqrt{3}} \]