Solve for \( x \) in the following equation: \( e^{x}-e^{-x}=3 \)
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To solve the equation \( e^{x} - e^{-x} = 3 \), we can start by letting \( y = e^{x} \). This means that \( e^{-x} = \frac{1}{y} \), and we can rewrite the equation as: \[ y - \frac{1}{y} = 3 \] Multiplying through by \( y \) to eliminate the fraction, we get: \[ y^2 - 1 = 3y \] Rearranging the terms gives us a quadratic equation: \[ y^2 - 3y - 1 = 0 \] Now, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -3, c = -1 \): \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2} \] This gives us two possible values for \( y \): \[ y_1 = \frac{3 + \sqrt{13}}{2}, \quad y_2 = \frac{3 - \sqrt{13}}{2} \] Since \( y = e^{x} \) must be positive, we discard \( y_2 \) because \( \sqrt{13} \) is greater than 3, making \( y_2 \) negative. Thus, we only consider: \[ y = \frac{3 + \sqrt{13}}{2} \] Now, to find \( x \), we take the natural logarithm: \[ x = \ln\left(\frac{3 + \sqrt{13}}{2}\right) \] So, the solution is: \[ x = \ln\left(\frac{3 + \sqrt{13}}{2}\right) \]
