`f(x) = (2x)/(log x)` is increasing in

Options

A. `(0, 1)`
B. `(1, e)`
C. `(e, oo)`
D. `(-oo, e)`

Explanation

To determine the intervals where the function \( f(x) = \frac{2x}{\log x} \) is increasing, we need to follow these steps:

### Step 1: Find the Derivative
We will use the quotient rule to find the derivative \( f'(x) \). The quotient rule states that if \( f(x) = \frac{u}{v} \), then:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Here, \( u = 2x \) and \( v = \log x \).

Calculating the derivatives:
- \( u' = 2 \)
- \( v' = \frac{1}{x} \)

Now applying the quotient rule:

\[ f'(x) = \frac{(2)(\log x) - (2x)\left(\frac{1}{x}\right)}{(\log x)^2} \]

This simplifies to:

\[ f'(x) = \frac{2 \log x - 2}{(\log x)^2} \]

### Step 2: Set the Derivative Greater than Zero
To find where the function is increasing, we need to determine where \( f'(x) > 0 \):

\[ \frac{2(\log x - 1)}{(\log x)^2} > 0 \]

Since \( (\log x)^2 \) is always positive for \( x > 1 \), we only need to consider the numerator:

\[ 2(\log x - 1) > 0 \implies \log x - 1 > 0 \implies \log x > 1 \]

### Step 3: Solve the Inequality
The inequality \( \log x > 1 \) can be solved by exponentiating both sides:

\[ x > e^1 \implies x > e \]

### Conclusion
Thus, the function \( f(x) = \frac{2x}{\log x} \) is increasing for:

\[ x \in (e, \infty) \]