Find the sum roots of `(e^(2x)-4)(6e^(2x)-5e^x+1)=0`

Options

A. `ln 6`
B. `ln 3`
C. `-ln 3`
D. `ln 2`

Answer

To solve the equation \((e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0\) and find the sum of its roots, we can follow these steps:

### Step 1: Set up the equation
The given equation is:
\[ (e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0 \]

### Step 2: Solve each factor separately
We can solve the equation by setting each factor to zero.

1. **First Factor**:
\[ e^{2x} - 4 = 0 \]
This gives:
\[ e^{2x} = 4 \]
Taking the natural logarithm on both sides:
\[ 2x = \ln(4) \implies x = \frac{\ln(4)}{2} = \ln(2) \]

2. **Second Factor**:
\[ 6e^{2x} - 5e^x + 1 = 0 \]
Let \(t = e^x\), then \(e^{2x} = t^2\). The equation becomes:
\[ 6t^2 - 5t + 1 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} \]
Simplifying:
\[ t = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12} \]
Thus, we have:
\[ t_1 = \frac{6}{12} = \frac{1}{2}, \quad t_2 = \frac{4}{12} = \frac{1}{3} \]

### Step 3: Find the corresponding values of \(x\)
Now we convert back from \(t\) to \(x\):
- For \(t_1 = \frac{1}{2}\):
\[ e^x = \frac{1}{2} \implies x = \ln\left(\frac{1}{2}\right) = -\ln(2) \]
- For \(t_2 = \frac{1}{3}\):
\[ e^x = \frac{1}{3} \implies x = \ln\left(\frac{1}{3}\right) = -\ln(3) \]

### Step 4: Sum of the roots
Now we have the roots:
1. \(x_1 = \ln(2)\)
2. \(x_2 = -\ln(2)\)
3. \(x_3 = -\ln(3)\)

The sum of the roots is:
\[ \text{Sum} = \ln(2) + (-\ln(2)) + (-\ln(3)) = -\ln(3) \]

### Final Answer
The sum of the roots is:
\[ -\ln(3) \]