The number of real solutions of `1+|e^x-1|=e^x(e^x-2)`

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To solve the equation \( 1 + |e^x - 1| = e^x(e^x - 2) \), we will analyze both sides of the equation step by step.

### Step 1: Analyze the absolute value
The expression \( |e^x - 1| \) can be split into two cases based on the value of \( e^x \):
1. Case 1: \( e^x - 1 \geq 0 \) (i.e., \( e^x \geq 1 \) or \( x \geq 0 \))
- In this case, \( |e^x - 1| = e^x - 1 \).
- The equation becomes:
\[ 1 + (e^x - 1) = e^x(e^x - 2) \]
Simplifying this gives:
\[ e^x = e^x(e^x - 2) \]
Rearranging leads to:
\[ e^x - e^{2x} + 2e^x = 0 \implies e^{2x} - 3e^x = 0 \]

2. Case 2: \( e^x - 1 < 0 \) (i.e., \( e^x < 1 \) or \( x < 0 \))
- In this case, \( |e^x - 1| = -(e^x - 1) = 1 - e^x \).
- The equation becomes:
\[ 1 + (1 - e^x) = e^x(e^x - 2) \]
Simplifying this gives:
\[ 2 - e^x = e^{2x} - 2e^x \]
Rearranging leads to:
\[ e^{2x} - e^x - 2 = 0 \]

### Step 2: Solve the equations
**For Case 1:**
From \( e^{2x} - 3e^x = 0 \):
- Factor out \( e^x \):
\[ e^x(e^x - 3) = 0 \]
- This gives us two solutions:
1. \( e^x = 0 \) (not possible since \( e^x > 0 \))
2. \( e^x = 3 \) which implies \( x = \log 3 \).

**For Case 2:**
From \( e^{2x} - e^x - 2 = 0 \):
- Let \( y = e^x \), then we have:
\[ y^2 - y - 2 = 0 \]
- Factoring gives:
\[ (y - 2)(y + 1) = 0 \]
- This gives us two solutions:
1. \( y = 2 \) which implies \( e^x = 2 \) or \( x = \log 2 \).
2. \( y = -1 \) (not possible since \( e^x > 0 \)).

### Step 3: Summary of solutions
The valid solutions from both cases are:
1. \( x = \log 3 \) from Case 1.
2. \( x = \log 2 \) from Case 2.

### Conclusion
Thus, the number of real solutions to the equation \( 1 + |e^x - 1| = e^x(e^x - 2) \) is **2**.