If the greatest value of `y= x/log x ` on `[e, e^(3)]` is `u` then `u` is equal to (given e= 2.71)

Answer

To find the greatest value of the function \( y = \frac{x}{\log x} \) on the interval \([e, e^3]\), we will follow these steps:

### Step 1: Find the derivative
To find the critical points, we need to differentiate the function \( y \).

Using the quotient rule:
\[ y = \frac{x}{\log x} \]
Let \( u = x \) and \( v = \log x \). Then,
\[ y' = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{1}{x} \).

Substituting these into the quotient rule gives:
\[ y' = \frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} \]

### Step 2: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ \frac{\log x - 1}{(\log x)^2} = 0 \]
This implies:
\[ \log x - 1 = 0 \implies \log x = 1 \implies x = e \]

### Step 3: Evaluate the function at the endpoints and critical points
Now we will evaluate \( y \) at the endpoints \( e \) and \( e^3 \), as well as at the critical point \( e \).

1. **At \( x = e \)**:
\[ y = \frac{e}{\log e} = \frac{e}{1} = e \approx 2.71 \]

2. **At \( x = e^3 \)**:
\[ y = \frac{e^3}{\log(e^3)} = \frac{e^3}{3 \log e} = \frac{e^3}{3} \approx \frac{(2.71)^3}{3} \]
Calculating \( (2.71)^3 \):
\[ (2.71)^3 \approx 19.927 \implies y \approx \frac{19.927}{3} \approx 6.6423 \]

### Step 4: Compare the values
Now we compare the values:
- At \( x = e \), \( y \approx 2.71 \)
- At \( x = e^3 \), \( y \approx 6.6423 \)

The greatest value \( u \) is:
\[ u \approx 6.6423 \]

### Conclusion
Thus, the greatest value of \( y = \frac{x}{\log x} \) on the interval \([e, e^3]\) is approximately \( 6.6423 \).

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