The sum of all the real roots of the equation `(e^(2x)-4)(6e^(2x)-5e^(x)+1)=0` is

Options

A. `log_(e)3`
B. `-log_(e)3`
C. `log_(e)6`
D. `-log_(e)6`

Answer

To solve the equation \((e^{2x}-4)(6e^{2x}-5e^{x}+1)=0\), we will break it down into two separate equations and solve for \(x\).

### Step 1: Solve \(e^{2x} - 4 = 0\)

1. Set the first factor equal to zero:
\[ e^{2x} - 4 = 0 \]
2. Rearranging gives:
\[ e^{2x} = 4 \]
3. Taking the natural logarithm of both sides:
\[ 2x = \ln(4) \]
4. Dividing by 2:
\[ x = \frac{\ln(4)}{2} = \ln(2) \]

### Step 2: Solve \(6e^{2x} - 5e^{x} + 1 = 0\)

1. Let \(y = e^{x}\). Then \(e^{2x} = y^2\), and we can rewrite the equation:
\[ 6y^2 - 5y + 1 = 0 \]
2. Now, we will use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = -5\), and \(c = 1\):
\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} \]
\[ y = \frac{5 \pm \sqrt{25 - 24}}{12} \]
\[ y = \frac{5 \pm 1}{12} \]
3. This gives us two possible values for \(y\):
\[ y_1 = \frac{6}{12} = \frac{1}{2}, \quad y_2 = \frac{4}{12} = \frac{1}{3} \]
4. Now, we convert back to \(x\):
- For \(y_1 = \frac{1}{2}\):
\[ e^{x} = \frac{1}{2} \implies x = \ln\left(\frac{1}{2}\right) = -\ln(2) \]
- For \(y_2 = \frac{1}{3}\):
\[ e^{x} = \frac{1}{3} \implies x = \ln\left(\frac{1}{3}\right) = -\ln(3) \]

### Step 3: Sum of all real roots

Now we have three real roots:
1. \(x_1 = \ln(2)\)
2. \(x_2 = -\ln(2)\)
3. \(x_3 = -\ln(3)\)

The sum of all real roots is:
\[ \text{Sum} = \ln(2) + (-\ln(2)) + (-\ln(3)) = 0 - \ln(3) = -\ln(3) \]

### Final Answer:
The sum of all the real roots of the equation is \(-\ln(3)\).