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

Answer

To solve the given problem, we need to find the number of points of intersection for the two equations:

1. \( e^x = x^2 \)
2. \( \log_e x = x \)

### Step 1: Analyze the first equation \( e^x = x^2 \)

We will analyze the functions \( y = e^x \) and \( y = x^2 \).

- The function \( y = e^x \) is an exponential function that increases rapidly and is always positive for all real \( x \).
- The function \( y = x^2 \) is a quadratic function that opens upwards and has its vertex at the origin (0,0).

To find the points of intersection, we can sketch the graphs of both functions:

- The graph of \( y = e^x \) starts at (0,1) and rises steeply as \( x \) increases.
- The graph of \( y = x^2 \) starts at (0,0) and rises more slowly compared to \( e^x \).

By observing the graphs, we can see that they intersect at two points. Thus, we conclude that:

**Number of intersection points for \( e^x = x^2 \) is 2.**

### Step 2: Analyze the second equation \( \log_e x = x \)

Now we will analyze the functions \( y = \log_e x \) and \( y = x \).

- The function \( y = \log_e x \) is defined for \( x > 0 \) and increases slowly. It approaches negative infinity as \( x \) approaches 0 and passes through the point (1,0).
- The function \( y = x \) is a straight line with a slope of 1 that passes through the origin.

To find the points of intersection, we can sketch the graphs of both functions:

- The graph of \( y = \log_e x \) starts from negative infinity at \( x = 0 \) and increases, crossing the x-axis at (1,0).
- The graph of \( y = x \) is a straight line that will eventually rise above \( \log_e x \) as \( x \) increases.

By observing the graphs, we can see that there are no points where \( \log_e x \) intersects \( y = x \). Thus, we conclude that:

**Number of intersection points for \( \log_e x = x \) is 0.**

### Final Answer

1. For \( e^x = x^2 \): 2 intersection points
2. For \( \log_e x = x \): 0 intersection points

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