Find the number of points of intersection
`(i) e^(x)=x^(2)` `(ii) log_(e)x=x`

Answer

To find the number of points of intersection for the given equations, we will analyze each equation step by step.

### Part (i): Finding the number of points of intersection for \( e^x = x^2 \)

1. **Understanding the Functions**: We have two functions:
- \( f(x) = e^x \)
- \( g(x) = x^2 \)

2. **Graphing the Functions**:
- The graph of \( e^x \) is an exponential curve that starts from \( (0, 1) \) and increases rapidly as \( x \) increases.
- The graph of \( x^2 \) is a parabola that opens upwards, starting from the origin \( (0, 0) \).

3. **Finding Points of Intersection**:
- We need to find where these two graphs intersect. This means we are looking for values of \( x \) such that \( e^x = x^2 \).
- By observing the graphs, we can see that they intersect at two points: one in the negative \( x \) region and one in the positive \( x \) region.

4. **Conclusion for Part (i)**:
- Therefore, the number of points of intersection for \( e^x = x^2 \) is **2**.

### Part (ii): Finding the number of points of intersection for \( \log_e x = x \)

1. **Understanding the Functions**: We have:
- \( f(x) = \log_e x \) (which is the natural logarithm of \( x \))
- \( g(x) = x \)

2. **Graphing the Functions**:
- The graph of \( \log_e x \) is defined for \( x > 0 \) and increases slowly, approaching negative infinity as \( x \) approaches 0.
- The graph of \( y = x \) is a straight line passing through the origin with a slope of 1.

3. **Finding Points of Intersection**:
- We need to find where \( \log_e x = x \).
- By observing the graphs, we can see that the line \( y = x \) will intersect the curve \( y = \log_e x \) only once in the positive \( x \) region.

4. **Conclusion for Part (ii)**:
- Therefore, the number of points of intersection for \( \log_e x = x \) is **0**.

### Final Answer:
- For \( e^x = x^2 \): **2 points of intersection**
- For \( \log_e x = x \): **0 points of intersection**