The equation x logx =3-x

Options

A. has no root in (1, 3)
B. has exactly one root in (1, 3)
C. `x log x -(3-x) gt 0 `in [1,3]
D. x log x-(3-x) `lt 0` in [1,3]

Answer

To solve the equation \( x \log x = 3 - x \), we can rewrite it as:

\[ f(x) = x \log x + x - 3 = 0 \]

### Step 1: Define the function
Let \( f(x) = x \log x + x - 3 \).

### Step 2: Evaluate the function at specific points
We will evaluate \( f(x) \) at \( x = 1 \) and \( x = 3 \).

- For \( x = 1 \):
\[ f(1) = 1 \log 1 + 1 - 3 = 1 \cdot 0 + 1 - 3 = 1 - 3 = -2 \]

- For \( x = 3 \):
\[ f(3) = 3 \log 3 + 3 - 3 = 3 \log 3 + 0 = 3 \log 3 \]
Since \( \log 3 \) is positive, \( f(3) > 0 \).

### Step 3: Analyze the sign of the function
From our evaluations:
- \( f(1) = -2 < 0 \)
- \( f(3) = 3 \log 3 > 0 \)

Since \( f(1) < 0 \) and \( f(3) > 0 \), by the Intermediate Value Theorem, there must be at least one root in the interval \( (1, 3) \).

### Step 4: Determine the nature of the function
Now we need to check if there is exactly one root in the interval \( (1, 3) \). To do this, we can analyze the derivative of \( f(x) \).

### Step 5: Calculate the derivative
The derivative of \( f(x) \) is:
\[ f'(x) = \log x + 1 \]
This derivative is defined for \( x > 0 \) and is positive for \( x > 1 \) since \( \log x > 0 \) for \( x > 1 \).

### Step 6: Conclusion about the function
Since \( f'(x) > 0 \) for \( x > 1 \), the function \( f(x) \) is increasing in the interval \( (1, 3) \). Therefore, it can cross the x-axis only once in this interval.

### Final Answer
Thus, we conclude that the equation \( x \log x = 3 - x \) has exactly one root in the interval \( (1, 3) \).