Prove that the equation `x 2^x= 1` has at least one positive root which is less than unity.

Answer

To prove that the equation \( x \cdot 2^x = 1 \) has at least one positive root which is less than 1, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x \cdot 2^x - 1 \). We want to find the roots of \( f(x) = 0 \).

### Step 2: Evaluate the function at specific points
First, we evaluate \( f(x) \) at \( x = 0 \) and \( x = 1 \):
- \( f(0) = 0 \cdot 2^0 - 1 = 0 - 1 = -1 \)
- \( f(1) = 1 \cdot 2^1 - 1 = 2 - 1 = 1 \)

### Step 3: Analyze the function's behavior
Since \( f(0) = -1 \) and \( f(1) = 1 \), we see that:
- \( f(0) < 0 \)
- \( f(1) > 0 \)

### Step 4: Apply the Intermediate Value Theorem
Since \( f(x) \) is continuous (as it is a product of continuous functions), and it changes sign between \( x = 0 \) and \( x = 1 \), by the Intermediate Value Theorem, there exists at least one \( c \in (0, 1) \) such that \( f(c) = 0 \).

### Step 5: Conclusion
Thus, we conclude that the equation \( x \cdot 2^x = 1 \) has at least one positive root which is less than 1.

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