Key Concepts and Formulas

• Exponential-to-Polynomial Transformation: Substituting $e^x$ with a variable (e.g., $t$) to transform an exponential equation into a polynomial equation.
• Domain of Exponential Functions: $e^x > 0$ for all real numbers $x$. This is crucial for validating solutions after substitution.
• Logarithm Properties: The inverse relationship between exponential and logarithmic functions: $e^x = t \iff x = \ln t$ (for $t > 0$), and $\ln(a/b) = \ln a - \ln b$, $\ln(a^b) = b \ln a$, $\ln(1/a) = -\ln a$.

Step-by-Step Solution

1. Rewriting the Equation The given equation is: $(e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0$ We observe that $e^{2x} = (e^x)^2$. This suggests a substitution to simplify the equation into a polynomial form.

2. Substitution to Simplify Let $t = e^x$. This substitution converts the equation into an algebraic equation in $t$. Since $x$ is a real number, $t = e^x > 0$. Substituting $t$ into the equation, we get: $(t^2 - 4)(6t^2 - 5t + 1) = 0$

3. Solving the Polynomial Equation for $t$ The equation $(t^2 - 4)(6t^2 - 5t + 1) = 0$ is a product of two factors that equals zero. Thus, at least one of the factors must be zero.

• Solving $t^2 - 4 = 0$: This is a difference of squares: $t^2 - 4 = (t-2)(t+2) = 0$. This gives us two potential values for $t$: $t = 2 \quad \text{or} \quad t = -2$

• Solving $6t^2 - 5t + 1 = 0$: This is a quadratic equation that can be factored. We look for two numbers that multiply to $6 \times 1 = 6$ and add up to $-5$. These numbers are $-2$ and $-3$. $6t^2 - 3t - 2t + 1 = 0$ $3t(2t - 1) - 1(2t - 1) = 0$ $(3t - 1)(2t - 1) = 0$ This gives us two more potential values for $t$: $t = \frac{1}{3} \quad \text{or} \quad t = \frac{1}{2}$

4. Identifying Valid Real Roots The potential values for $t$ are $2, -2, \frac{1}{2}, \frac{1}{3}$. Since $t = e^x > 0$, we must filter out any non-positive solutions for $t$.

• $t = 2$: This is positive, so it's a valid solution.
• $t = -2$: This is negative and is rejected.
• $t = \frac{1}{2}$: This is positive, so it's a valid solution.
• $t = \frac{1}{3}$: This is positive, so it's a valid solution.

Therefore, the valid values for $t$ that yield real roots for $x$ are $2, \frac{1}{2}, \text{ and } \frac{1}{3}$.

5. Finding the Real Roots $x$ Now, we convert these valid $t$ values back into $x$ values using the relationship $e^x = t$ and by applying the natural logarithm.

• For $t = 2$: $e^x = 2$ Taking the natural logarithm of both sides: $x = \ln 2$

• For $t = \frac{1}{2}$: $e^x = \frac{1}{2}$ Taking the natural logarithm of both sides: $x = \ln\left(\frac{1}{2}\right) = \ln(2^{-1}) = -\ln 2$

• For $t = \frac{1}{3}$: $e^x = \frac{1}{3}$ Taking the natural logarithm of both sides: $x = \ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln 3$

The set of all real roots for the original equation is $\{\ln 2, -\ln 2, -\ln 3\}$.

6. Calculating the Sum of Real Roots To find the sum of all real roots, we add the values: $\text{Sum} = (\ln 2) + (-\ln 2) + (-\ln 3)$ $\text{Sum} = \ln 2 - \ln 2 - \ln 3$ $\text{Sum} = -\ln 3$

Common Mistakes & Tips

• Crucial Tip: Always remember that $e^x > 0$ for all real values of $x$. Discard any non-positive solutions for $t$ after the substitution.
• Factoring Accuracy: Ensure accurate factoring of quadratic expressions to avoid incorrect potential roots.
• Logarithm Properties Application: Apply logarithm rules precisely. For instance, correctly applying $\ln(1/a) = -\ln a$ is essential.

Summary By transforming the exponential equation into a polynomial equation using $t = e^x$, solving for $t$, and filtering for valid positive values of $t$, we identified the real roots as $\ln 2$, $-\ln 2$, and $-\ln 3$. The sum of these roots is $-\ln 3$.

Final Answer The final answer is \boxed{{\log _e}3}, which corresponds to option (A).