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 rewrite this in terms of hyperbolic functions. The left-hand side is equivalent to \( 2\sinh(x) \), where \( \sinh(x) \) is the hyperbolic sine function. Thus, we have: \[ 2\sinh(x) = 3 \] Dividing both sides by 2 gives: \[ \sinh(x) = \frac{3}{2} \] Now, we can solve for \( x \) using the inverse hyperbolic sine function: \[ x = \sinh^{-1}\left(\frac{3}{2}\right) \] Using the formula for the inverse hyperbolic sine function, we have: \[ x = \ln\left(\frac{3}{2} + \sqrt{\left(\frac{3}{2}\right)^2 + 1}\right) \] Calculating inside the square root: \[ \left(\frac{3}{2}\right)^2 + 1 = \frac{9}{4} + 1 = \frac{9}{4} + \frac{4}{4} = \frac{13}{4} \] Now we take the square root: \[ \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \] Therefore, the expression for \( x \) becomes: \[ x = \ln\left(\frac{3}{2} + \frac{\sqrt{13}}{2}\right) = \ln\left(\frac{3 + \sqrt{13}}{2}\right) \] So the final solution is: \[ x = \ln\left(\frac{3 + \sqrt{13}}{2}\right) \]