`e^(4x)+2e^(3x)-e^x-6=0` find the number of solutions

Answer

To solve the equation \( e^{4x} + 2e^{3x} - e^x - 6 = 0 \) and find the number of solutions, we can follow these steps:

### Step 1: Substitution
Let's make a substitution to simplify the equation. Let \( y = e^x \). Then, we can rewrite the equation as:
\[ y^4 + 2y^3 - y - 6 = 0 \]

### Step 2: Analyze the Function
Define the function:
\[ f(y) = y^4 + 2y^3 - y - 6 \]
We need to find the number of real roots of this polynomial function.

### Step 3: Evaluate the Function at Specific Points
First, let's evaluate the function at some specific points to understand its behavior.

1. **At \( y = 1 \)**:
\[ f(1) = 1^4 + 2 \cdot 1^3 - 1 - 6 = 1 + 2 - 1 - 6 = -4 \]

2. **At \( y = 2 \)**:
\[ f(2) = 2^4 + 2 \cdot 2^3 - 2 - 6 = 16 + 16 - 2 - 6 = 24 \]

3. **At \( y = 0 \)**:
\[ f(0) = 0^4 + 2 \cdot 0^3 - 0 - 6 = -6 \]

### Step 4: Intermediate Value Theorem
From the evaluations:
- \( f(1) = -4 \) (negative)
- \( f(2) = 24 \) (positive)

Since \( f(y) \) changes from negative to positive between \( y = 1 \) and \( y = 2 \), by the Intermediate Value Theorem, there is at least one root in the interval \( (1, 2) \).

### Step 5: Find the Derivative
Next, we find the derivative of \( f(y) \) to analyze the number of turning points:
\[ f'(y) = 4y^3 + 6y^2 - 1 \]

### Step 6: Analyze the Derivative
To find critical points, we can set \( f'(y) = 0 \):
\[ 4y^3 + 6y^2 - 1 = 0 \]
This cubic equation can have up to 3 real roots. The nature of the roots will determine the number of turning points and thus the number of solutions to \( f(y) = 0 \).

### Step 7: Behavior at Infinity
As \( y \to \infty \), \( f(y) \to \infty \) (since the leading term \( y^4 \) dominates). As \( y \to -\infty \), \( f(y) \to -\infty \).

### Step 8: Conclusion
Since \( f(y) \) is a continuous polynomial function, and we have established that:
- There is at least one root in \( (1, 2) \).
- The function approaches \( -\infty \) as \( y \to -\infty \) and \( +\infty \) as \( y \to +\infty \).

By analyzing the behavior of \( f(y) \) and its derivative, we can conclude that there are **three real roots** for the equation \( f(y) = 0 \).

### Final Answer
Thus, the number of solutions to the equation \( e^{4x} + 2e^{3x} - e^x - 6 = 0 \) is **3**.

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