The question in the image asks to:

1. Express tanh(2x) in terms of $e^{2x}$, and
2. Show that $\ln\left(\frac{1 + \text{tanh}(2x)}{1 - \text{tanh}(2x)}\right) = 4x$.

Solution:

Part (a): Express $\tanh(2x)$ in terms of $e^{2x}$

The hyperbolic tangent function is defined as: $\tanh(2x) = \frac{\sinh(2x)}{\cosh(2x)}$ where: $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}, \quad \cosh(2x) = \frac{e^{2x} + e^{-2x}}{2}.$

Thus: $\tanh(2x) = \frac{\frac{e^{2x} - e^{-2x}}{2}}{\frac{e^{2x} + e^{-2x}}{2}} = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}.$

Part (b): Show $\ln\left(\frac{1 + \tanh(2x)}{1 - \tanh(2x)}\right) = 4x$

Using the expression for $\tanh(2x)$: $\tanh(2x) = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}},$ the numerator $1 + \tanh(2x)$ becomes: $1 + \tanh(2x) = 1 + \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \frac{(e^{2x} + e^{-2x}) + (e^{2x} - e^{-2x})}{e^{2x} + e^{-2x}} = \frac{2e^{2x}}{e^{2x} + e^{-2x}}.$

Similarly, the denominator $1 - \tanh(2x)$ becomes: $1 - \tanh(2x) = 1 - \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \frac{(e^{2x} + e^{-2x}) - (e^{2x} - e^{-2x})}{e^{2x} + e^{-2x}} = \frac{2e^{-2x}}{e^{2x} + e^{-2x}}.$

The fraction $\frac{1 + \tanh(2x)}{1 - \tanh(2x)}$ is then: $\frac{1 + \tanh(2x)}{1 - \tanh(2x)} = \frac{\frac{2e^{2x}}{e^{2x} + e^{-2x}}}{\frac{2e^{-2x}}{e^{2x} + e^{-2x}}} = \frac{2e^{2x}}{2e^{-2x}} = e^{4x}.$

Taking the natural logarithm: $\ln\left(\frac{1 + \tanh(2x)}{1 - \tanh(2x)}\right) = \ln(e^{4x}) = 4x.$

This completes the proof.

Would you like additional details or have any questions? Here are 5 related questions for deeper exploration:

1. What are the definitions of other hyperbolic functions, such as $\sinh(x)$ and $\cosh(x)$?
2. How does $\tanh(x)$ behave asymptotically as $x \to \infty$?
3. Can you derive similar logarithmic identities for other hyperbolic functions?
4. How can hyperbolic functions be applied to solve real-world problems in physics or engineering?
5. How does the relationship between $\tanh(2x)$ and $e^{2x}$ connect to complex exponentials?

Tip: Always verify logarithmic identities by rewriting the terms into their exponential equivalents—it often simplifies the work.