CrazCo Posted September 30, 2009 Share Posted September 30, 2009 e^x+e^-x=3 i tried xlne+-xlne=ln3 x(1)+-x(1) = ln3 x-x = ln3 but that doesn't work btw the answer is +- 0.96 Link to comment Share on other sites More sharing options...
Bignose Posted September 30, 2009 Share Posted September 30, 2009 have you learned about the hyperbolic trig functions yet? Link to comment Share on other sites More sharing options...
CrazCo Posted September 30, 2009 Author Share Posted September 30, 2009 no :S Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted September 30, 2009 Share Posted September 30, 2009 If you have this: [math]e^x + e^{-x} = 3[/math] taking the natural log of both sides gives you this: [math]\ln (e^x + e^{-x}) = \ln 3[/math] not what you wrote above. See what you can do using the properties of logarithms to tease x out of there. Link to comment Share on other sites More sharing options...
CrazCo Posted September 30, 2009 Author Share Posted September 30, 2009 tbh i have never seen anything like that. Link to comment Share on other sites More sharing options...
D H Posted September 30, 2009 Share Posted September 30, 2009 [math]\ln (e^x + e^{-x}) = \ln 3[/math] See what you can do using the properties of logarithms to tease x out of there. There is no teasing that x out of there. Bignose suggestion to look into the hyperbolic functions is exactly right. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted September 30, 2009 Share Posted September 30, 2009 Indeed. I just realized the method I had done quickly before breaks a few rules as well... but it does give me an answer of 0.97, which is confusingly close. Link to comment Share on other sites More sharing options...
ed84c Posted September 30, 2009 Share Posted September 30, 2009 My suggestion would be first to multiply through by e^x and move everything to one side (ie. you get ...... = 0). Then take a leap of faith as this this is a very similar function to solve to some you have been solving for years. (Hint: The answer to these is often = +/- .... likeyou have there) Link to comment Share on other sites More sharing options...
D H Posted September 30, 2009 Share Posted September 30, 2009 My suggestion would be first to multiply through by e^x and move everything to one side (ie. you get ...... = 0). Then take a leap of faith as this this is a very similar function to solve to some you have been solving for years. (Hint: The answer to these is often = +/- .... likeyou have there) Yep. That will do, also, and that is probably what is wanted for this (what appears to be homework) problem. Of course saying acosh(1.5) is much easier. Link to comment Share on other sites More sharing options...
ed84c Posted September 30, 2009 Share Posted September 30, 2009 Yup true. And I guess even if homework Qs arelooking for use of hyperbolics, it could be good practice to do it the long way round as you end up effectively deriving the equation for acosh, Link to comment Share on other sites More sharing options...
D H Posted October 1, 2009 Share Posted October 1, 2009 The hint ed84c gave is a bit cryptic (and appropriately so). Several hours have passed. A bit less cryptically, the idea is the solve for [math]e^x[/math]. Set [math]u=e^x[/math]. The equation in the original post becomes [math]u+1/u=3[/math]. Now multiply both sides by u. What kind of equation results from this? Link to comment Share on other sites More sharing options...
CrazCo Posted October 1, 2009 Author Share Posted October 1, 2009 LIKE THIS? e^x + 1/e^x = 3 e^2x -3(e^x)+1=0 Assume A = e^x A^2 - 3A + 1 = 0 Quadratic Formula 3+- sqrt 9-4/2 = A 3+- sqrt5/2 = A Replace 3+sqrt5/2 = e^x x = ln(3+sqrt5/2) = .96 X = ln(3-sqrt5/2) = -.96 im proud i got it!!!! Link to comment Share on other sites More sharing options...
D H Posted October 1, 2009 Share Posted October 1, 2009 There you go! Link to comment Share on other sites More sharing options...
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