The least value of `2log_(100)a-log_(a)(.0001),agt1` is

Options

A. 2
B. 3
C. 4
D. None

Answer

To find the least value of the expression \( 2\log_{100} a - \log_{a}(0.0001) \) for \( a > 1 \), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
We start with the expression:
\[ 2\log_{100} a - \log_{a}(0.0001) \]
Using the change of base formula, we can rewrite \( \log_{100} a \) as:
\[ \log_{100} a = \frac{\log_{10} a}{\log_{10} 100} = \frac{\log_{10} a}{2} \]
Thus, we have:
\[ 2\log_{100} a = 2 \cdot \frac{\log_{10} a}{2} = \log_{10} a \]

Now, we rewrite \( \log_{a}(0.0001) \):
\[ \log_{a}(0.0001) = \frac{\log_{10}(0.0001)}{\log_{10} a} = \frac{-4}{\log_{10} a} \]
because \( 0.0001 = 10^{-4} \).

### Step 2: Substitute back into the expression
Now, substituting these back into our original expression, we get:
\[ \log_{10} a - \left(-\frac{4}{\log_{10} a}\right) = \log_{10} a + \frac{4}{\log_{10} a} \]

### Step 3: Let \( x = \log_{10} a \)
Let \( x = \log_{10} a \). Then, our expression simplifies to:
\[ f(x) = x + \frac{4}{x} \]
where \( x > 0 \) since \( a > 1 \).

### Step 4: Find the minimum value of \( f(x) \)
To find the minimum value of \( f(x) \), we can use calculus. We take the derivative of \( f(x) \):
\[ f'(x) = 1 - \frac{4}{x^2} \]
Setting the derivative equal to zero to find critical points:
\[ 1 - \frac{4}{x^2} = 0 \implies \frac{4}{x^2} = 1 \implies x^2 = 4 \implies x = 2 \quad (\text{since } x > 0) \]

### Step 5: Verify it's a minimum
Now we check the second derivative:
\[ f''(x) = \frac{8}{x^3} \]
Since \( f''(x) > 0 \) for \( x > 0 \), \( f(x) \) has a local minimum at \( x = 2 \).

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

### Conclusion
Thus, the least value of \( 2\log_{100} a - \log_{a}(0.0001) \) for \( a > 1 \) is:
\[ \boxed{4} \]