To solve the equation \((e^{2x} – 4)(6e^{2x} – 5e^x + 1) = 0\), we need to find the values of \(x\) that satisfy it. Let's break down this equation and solve it step-by-step.
We have a product of two expressions set to zero. This implies that either of the expressions must equal zero. Hence, we can set up two equations:
\(e^{2x} - 4 = 0\)
\(6e^{2x} - 5e^x + 1 = 0\)
Let's solve the first equation \(e^{2x} - 4 = 0\):
Adding 4 to both sides gives \(e^{2x} = 4\).
Taking the natural logarithm on both sides: \(\ln(e^{2x}) = \ln(4)\).
This simplifies to \(2x = \ln(4)\), leading to \(x = \frac{\ln(4)}{2} = \ln(2)\).
Thus, one of the roots is \(x = \ln(2)\).
Now consider the second equation \(6e^{2x} - 5e^x + 1 = 0\):
Let \(y = e^x\). Then, \(e^{2x} = y^2\), reducing the equation to:
\[6y^2 - 5y + 1 = 0\]
This is a quadratic equation in \(y\). Applying the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get:
\[y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6}\]
\[ = \frac{5 \pm \sqrt{25 - 24}}{12}\]
\[ = \frac{5 \pm 1}{12}\]
The solutions for \(y\) are \(\frac{6}{12} = \frac{1}{2}\) and \(\frac{4}{12} = \frac{1}{3}\).
Reverting back to \(x\), \(e^x = \frac{1}{2}\) or \(e^x = \frac{1}{3}\).
Solving these gives:
If \(e^x = \frac{1}{2}\), then \(x = \ln(\frac{1}{2}) = -\ln(2)\).
If \(e^x = \frac{1}{3}\), then \(x = \ln(\frac{1}{3}) = -\ln(3)\).
Thus, the possible real roots are \(x = \ln(2)\), \(x = -\ln(2)\), and \(x = -\ln(3)\).
To find the sum of all real roots, we compute:
\[ \ln(2) + (-\ln(2)) + (-\ln(3)) = 0 - \ln(3) = -\ln(3) \]
Therefore, the sum of all the real roots is \(-\ln(3)\), matching the option \(-\log\;e^3\).
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