The number of real roots of equation `log_(e )x + ex = 0` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of real roots of the equation \( \log_e x + e^x = 0 \), we can follow these steps:

### Step 1: Rewrite the Equation
The given equation is:
\[ \log_e x + e^x = 0 \]
We can rewrite this as:
\[ \log_e x = -e^x \]

### Step 2: Analyze the Functions
We need to analyze the two functions:
1. \( y_1 = \log_e x \)
2. \( y_2 = -e^x \)

### Step 3: Determine the Domain
- The function \( \log_e x \) is defined for \( x > 0 \).
- The function \( -e^x \) is defined for all real \( x \).

### Step 4: Graph the Functions
- The graph of \( y_1 = \log_e x \) starts from negative infinity as \( x \) approaches 0 and increases without bound as \( x \) increases.
- The graph of \( y_2 = -e^x \) is a decreasing exponential function that starts from \( y = -1 \) when \( x = 0 \) and approaches negative infinity as \( x \) increases.

### Step 5: Identify Intersections
To find the number of real roots, we need to determine how many times these two graphs intersect:
- The graph of \( y_1 \) (logarithmic) will intersect the graph of \( y_2 \) (exponential) at most once because:
- \( y_1 \) increases without bound.
- \( y_2 \) decreases without bound.

### Step 6: Conclusion
Since \( y_1 \) is increasing and \( y_2 \) is decreasing, they can intersect at only one point. Therefore, the number of real roots of the equation \( \log_e x + e^x = 0 \) is:
\[ \text{Number of real roots} = 1 \]

### Final Answer
The number of real roots of the equation is **1**.

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