(b) Using the method of the equation [tex]$\frac{1+\tanh (x)}{1-\tanh (x)}=e^{2 x}$[/tex], with [tex]$x$[/tex] replaced by [tex]$y$[/tex]. Let [tex]$y=\tanh ^{-1}(x)$[/tex]. Then [tex]$x=$[/tex] ______ [tex]$\tanh (y)$[/tex], so we have the following.
[tex]\begin{aligned} e^{2 y} & =\frac{1+\tanh (y)}{1-\tanh (y)}=\frac{1+x}{1-x} \\ \Rightarrow \quad 2 y & =\ln \left(\frac{1+x}{\square}\right) \\ \Rightarrow \quad y & =\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \end{aligned}[/tex]
Answer (2)
Given y = tanh − 1 ( x ) , then x = tanh ( y ) .
Using the equation e 2 y = 1 − t a n h ( y ) 1 + t a n h ( y ) = 1 − x 1 + x , we find 2 y = ln ( 1 − x 1 + x ) .
Therefore, the missing term in 2 y = ln ( □ 1 + x ) is 1 − x .
Thus, x = tanh ( y ) and the missing term is 1 − x .
Explanation
Problem Setup We are given the equation 1 − t a n h ( x ) 1 + t a n h ( x ) = e 2 x , and we are asked to use this equation with x replaced by y to find the missing terms in the given equations.
Finding the first missing term Let y = tanh − 1 ( x ) . Then, by the definition of the inverse hyperbolic tangent function, we have x = tanh ( y ) .
Finding the second missing term We are given that e 2 y = 1 − t a n h ( y ) 1 + t a n h ( y ) = 1 − x 1 + x . Taking the natural logarithm of both sides, we get 2 y = ln ( 1 − x 1 + x ) . Therefore, the missing term in 2 y = ln ( □ 1 + x ) is 1 − x .
Final Answer Thus, we have x = tanh ( y ) and 2 y = ln ( 1 − x 1 + x ) , which implies y = 2 1 ln ( 1 − x 1 + x ) .
Examples
Imagine you're designing a navigation system that needs to calculate distances on a curved surface, like the Earth. The inverse hyperbolic tangent function, much like the one in this problem, can be used to map distances on a sphere to a flat plane, allowing for simpler calculations. By understanding how to manipulate and solve equations involving hyperbolic functions, you can create more accurate and efficient navigation tools.
Using the relationship y = tanh − 1 ( x ) , we find that x = tanh ( y ) . The missing term in the equation is 1 − x . Overall, this demonstrates how inverse hyperbolic functions can interrelate with exponential forms.
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