\( \left. \begin{array} { l l } { 5.3 } & { \frac { e ^ { x } - e ^ { - x } } { e ^ { x } + e ^ { - x } } = \frac { 1 } { 2 e } } \\ { 5.4 } & { e ^ { 2 \ln ( 2 x - 3 ) } = 3 x } \end{array} \right. \)
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To solve the equation \( \frac { e ^ { x } - e ^ { - x } } { e ^ { x } + e ^ { - x } } = \frac { 1 } { 2 e } \), recognize that this is the hyperbolic tangent function, \( \tanh(x) \). You can set \( \tanh(x) = \frac{1}{2e} \) and solve for \( x \). In the second equation \( e^{2 \ln(2x - 3)} = 3x \), use properties of logarithms to simplify: \( e^{\ln((2x - 3)^2)} = 3x \). This becomes \( (2x - 3)^2 = 3x \). Expanding and rearranging gives a quadratic equation to solve for \( x \).