Solve Exponential Function x for y: x = (e^y-e^{-y})/2

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Discussion Overview

The discussion revolves around solving the equation x = (e^y - e^{-y})/2 for y in terms of x. The scope includes mathematical reasoning and problem-solving techniques related to exponential functions.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • The original poster (OP) seeks assistance in solving the equation for y.
  • One participant suggests that the equation can be viewed as a quadratic in e^y, indicating that it should be solvable.
  • Another participant provides a detailed manipulation of the equation, leading to a solution involving logarithms.
  • A later reply emphasizes the importance of providing hints rather than full solutions to encourage independent problem-solving.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the equation, but there is a disagreement regarding the extent of help that should be provided, with some advocating for hints only.

Contextual Notes

Some assumptions about the familiarity with logarithmic and exponential functions may be implicit in the discussion. The steps leading to the solution involve transformations that may not be fully detailed.

Who May Find This Useful

Students and individuals interested in solving exponential equations, particularly in a homework context, may find this discussion beneficial.

expscv
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solve [tex]x =(e^y-e^{-y})/2[/tex] for y in terms of x

i kinda forgot how to do it~~~sign

any help appreciated

thx


could anyone move this to homework k-12 section thanks

sorry about this
 
Last edited:
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It's a quadratic in e^y, albeit disguised. You'd have no problem solving

2x= t - 1/t
 
[tex]2x=e^{y}-e^{-y}[/tex]
[tex]2x\cdot e^{y}=e^{2y}-1[/tex]
[tex]e^{y}=\frac{2x\pm\sqrt{4x^{2}+4}}{2}[/tex]
[tex]y=ln{(x\pm\sqrt{x^{2}+1})}[/tex]
 
The idea was that the OP would use Matt's suggestion and make the calculations on his/her own.
Please,do not provide full answers,once the 'HINTS' have been given...

Daniel.
 
cool thanks for the hint and the answer
 

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